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Mole Calculations - Practice Questions (38)

Question 1: 1. It is known that when $\mathrm { Cl } _ { 2 }$ is passed into an appropriate amount of KOH soluti...

1. It is known that when $\mathrm { Cl } _ { 2 }$ is passed into an appropriate amount of KOH solution, there may be $\mathrm { KCl } , \mathrm { KClO } , \mathrm { KClO } _ { 3 }$ in the product, and the value of the product depends on the temperature. When $n ( \mathrm { KOH } ) = a \mathrm {~mol}$ is $n ( \mathrm { KOH } ) = a \mathrm {~mol}$, the following statements are false

  • A. A. If the reaction is $= 11$ at a certain temperature, then the solution $=$
  • B. B. The amount of chlorine gas participating in the reaction is equal to $a \mathrm {~mol}$
  • C. C. The range of $n _ { e }$ for the amount of electrons transferred in a reaction by changing the temperature : $a \mathrm {~mol} \leq n _ { e } \leq a \mathrm {~mol}$
  • D. D. Changing the temperature, the maximum theoretical yield of $\mathrm { KClO } _ { 3 }$ in the product is $a \mathrm {~mol}$

Answer: D

Solution: 试题分析 :A. Let $n \left( \mathrm { ClO } ^ { - } \right) = 1 \mathrm {~mol}$, after the reaction $\mathrm { C } \left( \mathrm { Cl } ^ { - } \right) : \mathrm { C } \left( \mathrm { ClO } ^ { - } \right) = 11$, then $n \left( \mathrm { Cl } ^ { - } \right) = 11 \mathrm {~mol}$, the electron transfer is conserved, $5 \times \mathrm { n } \left( \mathrm { ClO } _ { 3 } { } ^ { - } \right) + 1 \times \mathrm { n } \left( \mathrm { ClO } ^ { - } \right) = 1 \times \mathrm { n } \left( \mathrm { Cl } ^ { - } \right)$, that is, $5 \times \mathrm { n } \left( \mathrm { ClO } _ { 3 } \right. \left. { } ^ { - } \right) + 1 \times 1 \mathrm {~mol} = 1 \times 11 \mathrm {~mol}$, the solution is $n \left( \mathrm { ClO } _ { 3 } { } ^ { - } \right) = 2 \mathrm {~mol}$, therefore the solution $=$, so A is correct. INLINE_FORMULA_5]], so the solution is $=$, so A is correct; B. From the conservation of Cl atoms, $2 n \left( \mathrm { Cl } _ { 2 } \right) = n ( \mathrm { KCl } ) + n ( \mathrm { KClO } ) + n \left( \mathrm { KClO } _ { 3 } \right)$, and from the conservation of potassium ions, $n ( \mathrm { KCl } ) + n ( \mathrm { KClO } ) + n \left( \mathrm { KClO } _ { 3 } \right) = n ( \mathrm { KOH } )$, so the amount of chlorine gas that takes part in the reaction is $= n ( \mathrm { KOH } ) = \mathrm { amol }$, so B is correct; C. oxidation products only $\mathrm { KClO } _ { 3 }$ when the most electrons are transferred, according to the conservation of electron transfer $\mathrm { n } ( \mathrm { KCl } ) = 5 \left( \mathrm { KClO } _ { 3 } \right)$, by the conservation of potassium ions: $n ( \mathrm { KCl } ) + n \left( \mathrm { KClO } _ { 3 } \right) = n ( \mathrm { KOH } )$, so $n \left( \mathrm { KClO } _ { 3 } \right) = n ( \mathrm { KOH } ) \times 1 / 6 =$ "$a$" $\mathrm { mol } \times 1 / 6$, the maximum amount of electrons transferred is: a mol $\times 5 \times 1 / 6 = \mathrm { a } \times 5 / 6 \mathrm {~mol}$, and the least amount of electrons is transferred when the only oxidation product is KClO, according to the conservation of electron transfer $\times 5 \times 1 / 6 = \mathrm { a } \times 5 / 6 \mathrm {~mol}$. Transfer conservation $n ( \mathrm { KCl } ) = n ( \mathrm { KClO } )$, according to the conservation of potassium ions: $n ( \mathrm { KCl } ) + n ( \mathrm { KClO } ) = \mathrm { n } ( \mathrm { KOH } )$, so $: \mathrm { n } ( \mathrm { KClO } ) = \mathrm { n } ( \mathrm { KOH } ) = \mathrm { a } \mathrm { mol }$, the smallest amount of electrons transferred $= \mathrm { a } \mathrm { mol } \times 1 = \mathrm { a } \mathrm { mol }$, then the amount of electrons transferred in the reaction [$= \mathrm { a } \mathrm { mol } \times 1 = \mathrm { a } \mathrm { mol }$. INLINE_FORMULA_21]] is in the range of a mol $\leq \mathrm { n } _ { \mathrm { e } } \leq \mathrm { a } \mathrm { mol }$, so C is correct; D. Oxidation product is only $\mathrm { KClO } _ { 3 }$ when the maximum amount of its material, according to the conservation of electron transfer $n ( \mathrm { KCl } ) = 5 \left( \mathrm { KClO } _ { 3 } \right)$, by the conservation of potassium ions $: n ( \mathrm { KCl } ) + n \left( \mathrm { KClO } _ { 3 } \right) = n ( \mathrm { KOH } )$, so $n _ { \text {最大 } } \left( \mathrm { KClO } _ { 3 } \right) = \frac { 1 } { 6 } n ( \mathrm { KOH } ) = \frac { 1 } { 6 }$ a mol, so D error.

Question 2: 2. $\mathrm { CS } _ { 2 }$ (liquid) can be burned in oxygen to form $\mathrm { CO } _ { 2 }$ and $\...

2. $\mathrm { CS } _ { 2 }$ (liquid) can be burned in oxygen to form $\mathrm { CO } _ { 2 }$ and $\mathrm { SO } _ { 2 }$ and $0.228 \mathrm {~g} \mathrm { CS } _ { 2 }$ is burned in 448 mL of standard condition $\mathrm { O } _ { 2 }$. Combine $0.228 \mathrm {~g} \mathrm { CS } _ { 2 }$ in 448 mL (standard conditions) of $\mathrm { O } _ { 2 }$, and return the combusted gas mixture to the standard condition volume of

  • A. A. 201.6 mL
  • B. B. 448 mL
  • C. C. 224 mL
  • D. D. 336 mL

Answer: B

Solution: Carbon disulfide $\left( \mathrm { CS } _ { 2 } \right)$ burns completely in oxygen to form $\mathrm { CO } _ { 2 }$ and $\mathrm { SO } _ { 2 }$, and the chemical equation for the reaction is: $\mathrm { CS } _ { 2 } + 3 \mathrm { O } _ { 2 } = \mathrm { CO } _ { 2 } + 2 \mathrm { SO } _ { 2 }$, carbon disulfide ($\mathrm { CS } _ { 2 }$) is a liquid. _4]], carbon disulfide ( $\mathrm { CS } _ { 2 }$) is a liquid, and 0.228 g of carbon disulfide has an amount of substance of 0.003 mol, and 0.009 mol of oxygen is needed for its combustion. The amount of oxygen in 448 mL of oxygen is 0.02 mol, so there is enough oxygen. The stoichiometric number of the gas before and after the reaction remains unchanged, and the total volume of the gas after combustion remains unchanged, so the volume of the gas after the reaction is 448 mL.

Question 3: 3. The following solutions contain the same amount of $100 \mathrm {~mL} 0.5 \mathrm {~mol} / \mathr...

3. The following solutions contain the same amount of $100 \mathrm {~mL} 0.5 \mathrm {~mol} / \mathrm { L } \mathrm { NaCl }$ as $\mathrm { Cl } ^ { - }$.

  • A. A. $100 \mathrm {~mL} 0.5 \mathrm {~mol} / \mathrm { L } \mathrm { MgCl } _ { 2 }$ solution
  • B. B. $200 \mathrm {~mL} 0.25 \mathrm {~mol} / \mathrm { L } \mathrm { CaCl } _ { 2 }$ solution
  • C. C. $50 \mathrm { ml } 1 \mathrm {~mol} / \mathrm { L } \mathrm { NaCl }$ solution
  • D. D. $25 \mathrm { ml } 2.0 \mathrm {~mol} / \mathrm { L } \mathrm { HCl }$ solution

Answer: B

Solution: A. The $100 \mathrm {~mL} 0.5 \mathrm {~mol} / \mathrm { L } \mathrm { MgCl } _ { 2 }$ solution contains $\mathrm { Cl } ^ { - }$ at a concentration of $1.0 \mathrm {~mol} / \mathrm { L }$; B. The $200 \mathrm {~mL} 0.25 \mathrm {~mol} / \mathrm { L } \mathrm { CaCl } _ { 2 }$ solution contains $\mathrm { Cl } ^ { \prime }$ in the amount of $0.5 \mathrm {~mol} / \mathrm { L }$, and B is correct; C. The $50 \mathrm { ml } 1 \mathrm {~mol} / \mathrm { L } \mathrm { NaCl }$ solution contains $\mathrm { Cl } ^ { - }$ at a concentration of $1 \mathrm {~mol} / \mathrm { L }$, and C is wrong; D. The $25 \mathrm { ml } 2.0 \mathrm {~mol} / \mathrm { L } \mathrm { HCl }$ solution contains $\mathrm { Cl } ^ { - }$ at a concentration of $2.0 \mathrm {~mol} / \mathrm { L }$, and D is wrong; The answer is B. [点睛]解答本题时要注意,题中题干中给出的溶液的体积是多余的条件,无需代入进行计算.

Question 4: 4. The correct statements about equal volumes of $\mathrm { N } _ { 2 } \mathrm { O }$ (gas) and $\m...

4. The correct statements about equal volumes of $\mathrm { N } _ { 2 } \mathrm { O }$ (gas) and $\mathrm { CO } _ { 2 }$ at the same temperature and pressure are (1) have the same mass (2) contain an equal number of carbon and nitrogen atoms (3) contain an equal number of molecules

  • A. A. (1) (2) (3)
  • B. B. (2) (3)
  • C. C. (1) (2)
  • D. D. (1) (3)

Answer: D

Solution: Two gases $\mathrm { N } _ { 2 } \mathrm { O }$ and $\mathrm { CO } _ { 2 }$ of equal volume at the same temperature and pressure have the same amount of substance and contain the same number of molecules, and the molar mass of both of them is $44 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 }$, so the masses of both of them are also equal; since the number of carbon atoms contained in the molecules of both are different from the number of nitrogen atoms, (2) is wrong. The number of carbon atoms and nitrogen atoms in the two molecules are different, so (2) is wrong, and the answer is D.

Question 5: 5. An atom X has a nuclear neutron number of $N$ and a mass number of $A$, and together with the ato...

5. An atom X has a nuclear neutron number of $N$ and a mass number of $A$, and together with the atom ${ } ^ { 2 } \mathrm { H }$ it forms a ${ } ^ { 2 } \mathrm { H } _ { \mathrm { m } } \mathrm { X }$ molecule. The quantity of protons contained in $a \mathrm {~g} ^ { 2 } \mathrm { H } _ { \mathrm { m } } \mathrm { X }$ is

  • A. A. $\frac { a } { A + 2 m } ( A - N + m ) \mathrm { mol }$
  • B. B. $\frac { a } { A + m } ( A - N + m ) \mathrm { mol }$
  • C. C. $\frac { a } { A + m } ( A - N ) \mathrm { mol }$
  • D. D. $\frac { a } { A } ( A - N ) \mathrm { mol }$

Answer: A

Solution: An atom X of an element has a mass number of $A$ and contains $N$ neutrons, then the number of protons is $( A - N )$, and each ${ } ^ { 2 } \mathrm { H } _ { \mathrm { m } } \mathrm { X }$ molecule contains $( A - N + m )$ of protons, and the amount of $\mathrm { ag } ^ { 2 } \mathrm { H } _ { \mathrm { m } } \mathrm { X }$ is: $\frac { a } { A + 2 m } \mathrm {~mol}$, then the amount of $\mathrm { ag } ^ { 2 } \mathrm { H } _ { \mathrm { m } } \mathrm { X }$ containing protons is [[]]. INLINE_FORMULA_8]] , and the answer is A .

Question 6: 7. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant.

7. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant.

  • A. A. Under high temperature, high pressure and catalyst conditions, sufficient amount of $\mathrm { N } _ { 2 }$ reacts with $3 \mathrm {~mol} \mathrm { H } _ { 2 }$ sufficiently to co-break the $\mathrm { H } - \mathrm { H }$ bond $3 \mathrm {~N} _ { \mathrm { A } }$.
  • B. B. 46 g Na reacts fully with $0.5 \mathrm {~L} 2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ hydrochloric acid and transfers electrons in the amount of $2 \mathrm {~N} _ { \mathrm { A } }$
  • C. C. 8.4 g solid $\mathrm { NaHCO } _ { 3 }$ containing cations $0.2 \mathrm {~N} _ { \mathrm { A } }$
  • D. D. In standard condition, $2.24 \mathrm {~L} \mathrm { CCl } _ { 4 }$ contains the number of covalent bonds as $0.4 \mathrm {~N} _ { \mathrm { A } }$

Answer: B

Solution: Test Question Analysis: A. The reaction between $\mathrm { N } _ { 2 }$ and $\mathrm { H } _ { 2 }$ is a reversible reaction, which can't be carried out completely, so A is wrong; B. The amount of 46 g Na is 2 mol, and there is 1 electron in the outermost layer of the sodium atom, which can be transferred to $0.5 \mathrm {~L} 2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ hydrochloric acid by reacting with $2 \mathrm {~N} _ { \mathrm { A } }$, so B is correct. C. The amount of $8.4 \mathrm {~g} \mathrm { NaHCO } _ { 3 }$ is 0.1 mol, and the amount of cations and anions in the solid is 1 mol, so C is wrong; D. Under standard conditions, $\mathrm { CCl } _ { 4 }$ is not a gas, so D is wrong; D. $\mathrm { CCl } _ { 4 }$ is not a gas at standard conditions. Avogadro(Avogadro)'s constant is a constant in the gas. [Notice] Avogadro's Constant: (1) whether the state of the substance is a gas; (2) whether the conditions are standard for a gas; (3) the difference of isotope atoms; (4) the number of ions in a reversible reaction or hydrolyzable salt; (5) the structure of the substance: e.g., $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ is a gas made up of $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ and $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ is a gas made up of [[]]. INLINE_FORMULA_7]] and $\mathrm { O } _ { 2 } { } ^ { 2 }$ instead of having $\mathrm { Na } ^ { + }$ and $\mathrm { O } ^ { 2 }$; $\mathrm { SiO } _ { 2 } , \mathrm { SiC }$ are atomic crystals, which have only atoms and no molecules, $\mathrm { SiO } _ { 2 }$ is an orthotetrahedral structure, $1 \mathrm { molSiO } _ { 2 }$ contains covalent bonds as $4 \mathrm {~N} _ { \mathrm { A } } , ~ 1 \mathrm { molP } _ { 4 }$ contains covalent bonds as $6 \mathrm {~N} _ { \mathrm { A } }$ and so on.

Question 7: 8. When comparing equal masses of ${ } ^ { \mathrm { O } _ { 2 } }$ and ${ } ^ { \mathrm { O } _ { 3...

8. When comparing equal masses of ${ } ^ { \mathrm { O } _ { 2 } }$ and ${ } ^ { \mathrm { O } _ { 3 } }$, the following statements are correct. $\begin{array} { l l l } \text {(1)molecular ratio is } 2 : 3 & \text {(2)molar ratio } 3 : 2 & \text {(3)oxygen atom ratio is } 1 : 1 \end{array}$ (4) The ratio of elemental oxygen masses is $3 : 2$ (5) The ratio of electron numbers is $2 : 3$

  • A. A. (1) (4)
  • B. B. (2) (3)
  • C. C. (4) (5)
  • D. D. (1) (2) (3) (4) (5)

Answer: B

Solution: The ratio of the amounts of $\mathrm { O } _ { 2 }$ and $\mathrm { O } _ { 3 }$ substances of equal mass $\frac { 1 } { 32 } : \frac { 1 } { 48 } = 3 : 2$; (1) The number of particles is proportional to the amount of matter, and the ratio of the number of molecules is $3 : 2$, which is wrong; (2) The ratio of the amount of matter $3 : 2$, is correct; (3) The ratio of the number of oxygen atoms is $( 3 \times 2 ) : ( 2 \times 3 ) = 1 : 1$ , which is correct; (4) The ratio of the number of oxygen atoms is $( 3 \times 2 ) : ( 2 \times 3 ) = 1 : 1$ and the ratio of the mass of elemental oxygen is $1 : 1$ , which is incorrect; (5) The ratio of the number of oxygen atoms is $( 3 \times 2 ) : ( 2 \times 3 ) = 1 : 1$ , then the ratio of the number of electrons is $1 : 1$ , which is incorrect;

Question 8: 9. In each of the following solutions, the concentration of chloride ions is equal to the concentrat...

9. In each of the following solutions, the concentration of chloride ions is equal to the concentration of chloride ions in the $50 \mathrm {~mL} 1 \mathrm {~mol} / \mathrm { L }$ solution of $\mathrm { AlCl } _ { 3 }$.

  • A. A. NaCl solution of $150 \mathrm {~mL} 1 \mathrm {~mol} / \mathrm { L }$.
  • B. B. $75 \mathrm {~mL} 1 \mathrm {~mol} / \mathrm { L }$ of $\mathrm { NH } _ { 4 } \mathrm { Cl }$ solution
  • C. C. KCl solution of $150 \mathrm {~mL} 3 \mathrm {~mol} / \mathrm { L }$.
  • D. D. $75 \mathrm {~mL} 2 \mathrm {~mol} / \mathrm { L }$ of $\mathrm { CaCl } _ { 2 }$ solution

Answer: C

Solution: A. A NaCl solution of $\mathrm { lmol } \cdot \mathrm { L } ^ { - 1 }$ with a concentration of chloride ions of $1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } , \mathrm {~A}$ does not meet the question; B. The concentration of chloride ions in a $\mathrm { NH } _ { 4 } \mathrm { Cl }$ solution of $\mathrm { NH } _ { 4 } \mathrm { Cl }$ is $1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$; C. The concentration of chloride ions in a KCl solution of $3 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ is $\mathrm { mol } \cdot \mathrm { L } ^ { - 1 } \times 1 = 3 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } , \mathrm { C }$; D. The concentration of chloride ions in $2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$'s $\mathrm { CaCl } _ { 2 }$ solution is $2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \times 2 = 4 \mathrm {~mol} / \mathrm { L } , \mathrm { D }$; C. The concentration of chloride ions in $2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \times 2 = 4 \mathrm {~mol} / \mathrm { L } , \mathrm { D }$'s $2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \times 2 = 4 \mathrm {~mol} / \mathrm { L } , \mathrm { D }$ solution is not consistent with the meaning of the question.

Question 9: 10. A student records the following data in the lab report, which is correct

10. A student records the following data in the lab report, which is correct

  • A. A. The pH of a solution is 3.5 when measured with a wide range of pH paper.
  • B. B. Measure 7.36 mL of hydrochloric acid with a 10 mL measuring cylinder.
  • C. C. Prepare 980 mL of solution in a 1000 mL volumetric flask.
  • D. D. Weigh 8.75 g of table salt on a pallet balance.

Answer: C

Solution: A. The pH of a solution measured on a wide range of pH paper and read against a colorimeter can only be recorded as a whole number value of 3 or 4, not 3.5, which is incorrect; B. The 10 mL measuring cylinder is accurate to 0.1 mL and measures 7.4 mL; C. To prepare a 980 mL solution, use a 1000 mL volumetric flask because there is no 980 mL volumetric flask; D. The pallet balance is accurate to 0.1 g, and 8.75 g of table salt should be weighed on the pallet balance and recorded as 8.8 g. D is incorrect; therefore, C is chosen.

Question 10: 11. In order to prevent the emergence of an epidemic after a disaster, epidemiologists disinfect wit...

11. In order to prevent the emergence of an epidemic after a disaster, epidemiologists disinfect with peroxyacetic acid ($\mathrm { C } _ { 2 } \mathrm { H } _ { 4 } \mathrm { O } _ { 3 }$) and sodium hypochlorite (NaClO). The following statements are correct

  • A. A. The valence of the element chlorine in sodium hypochlorite is - 1
  • B. B. Sodium hypochlorite is an oxide.
  • C. C. The relative molecular mass of peroxyacetic acid is 76
  • D. D. The mass ratio of $\mathrm { C } , \mathrm { H } , \mathrm { O }$ elements in peroxyacetic acid is $2 : 4 : 3$

Answer: C

Solution: A. The algebraic sum of the valences of the elements in a compound is zero. In sodium hypochlorite, the element sodium has a valence of +1 and the element oxygen has a valence of -2, so the valence of the element chlorine is +1, so A is wrong; B. There are two elements, one of which is a compound of oxygen is called an oxide, by the metal cation or ammonium ions and acid ions combined with the formation of compounds is called salt, sodium hypochlorite by the composition of the three elements, is not an oxide, sodium hypochlorite by the composition of sodium ions and hypochlorite ions, is a salt, so the B error; C. According to the molecular formula of peroxyacetic acid, relative molecular mass is numerically equal to molar mass $= 2 \times 12 + 4 \times 1 + 3 \times 16 = 76$, so C is correct; D. According to the molecular formula of peroxyacetic acid, the mass ratio of C, H and O elements in the molecule is $2 \times 12 : 4 \times 1 : 3 \times 16 = 6 : 1 : 12$, so D is wrong; Answer choice C.

Question 11: 12. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant. The following state...

12. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant. The following statement is correct

  • A. A. $2 \mathrm {~L} 0.5 \mathrm {~mol} / \mathrm { L }$ The number of $\mathrm { H } ^ { + }$ ions contained in a sulfite solution is $2 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. The number of $\mathrm { OH } ^ { - }$ ions ionized from water in a NaOH solution of $1 \mathrm { LpH } = 13$ at room temperature is $0.1 \mathrm {~N} _ { \mathrm { A } }$
  • C. C. The mass of copper dissolved at the anode during electrolytic refining of copper is 32 g for every $\mathrm { N } _ { \mathrm { A } }$ electron transferred
  • D. D. A closed container containing $0.1 \mathrm {~mol} \mathrm {~N} _ { 2 }$ and $0.3 \mathrm {~mol} \mathrm { H } _ { 2 }$ reacts fully under certain conditions and transfers fewer electrons than $0.6 \mathrm {~N} _ { \mathrm { A } }$

Answer: D

Solution: A. $2 \mathrm {~L} 0.5 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ sulfite solution contains fewer $\mathrm { H } ^ { + }$ ions than $2 \mathrm {~N} _ { \mathrm { A } }$, so it doesn't fit the question; B. In sodium hydroxide solution, hydroxide ions inhibit the ionization of water, and hydrogen ions in solution are water ionized, so the concentration of water ionized hydroxide ions $=$ hydrogen ions $= 1 \times 10 ^ { - 13 } \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ , so it does not meet the question; C. electrolysis refining copper, in addition to copper discharge on the anode, there are more active than copper metal discharge, so when the transfer of $\mathrm { N } _ { 4 }$ electrons, the anode of the mass reduction is less than 32 g, so does not meet the meaning of the question; D. The reaction of synthesizing nitrogen is reversible and cannot be carried out completely, so the number of electrons transferred is less than $0.6 \mathrm {~N} _ { 4 }$, so it is consistent with the meaning of the question. [Eyes]to answer the calculation of Avogadro's constant, pay attention to the following questions: (1) the state of aggregation of substances: the state of aggregation of substances is related to the condition in which they are located. (2) the conditions of use of the molar volume of gas: in non-standard conditions (such as room temperature, atmospheric pressure), 22.4 L of gas is not the amount of 1 mol; therefore, in non-standard conditions, can not be used to calculate the amount of gas molar volume of gas. (3) the composition of substances: the composition of substances in the number of molecules, the number of atoms or the number of ions in the accurate judgment is the key to answering these questions. (4) the number of electrons transferred in special chemical reactions.

Question 12: 13. $N _ { \mathrm { A } }$ is the value of Avogadro's constant, and the following statement is corr...

13. $N _ { \mathrm { A } }$ is the value of Avogadro's constant, and the following statement is correct

  • A. A. The number of electrons in 1 mol of hydroxyl is $9 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. 56 gFe reacts completely with a sufficient amount of nonmetallic monomers by heating, the number of electrons transferred is $2 N _ { \mathrm { A } }$
  • C. C. The total number of $0.1 \mathrm {~mol} { } ^ { \mathrm { NH } _ { 3 } }$ in the solution obtained by dissolving $\mathrm { NH } _ { 3 } \cdot \mathrm { H } _ { 2 } \mathrm { O } _ { \text {和 } } \mathrm { NH } _ { 4 } ^ { + }$ in water is equal to $0.1 \mathrm {~N} _ { \mathrm { A } }$.
  • D. D. In a zinc-copper primary cell, the theoretical number of electrons transferred when the negative electrode is reduced by 65 g is $N _ { \mathrm { A } }$

Answer: A

Solution: A. The hydroxyl group (- OH) has 8 electrons per oxygen atom and 1 electron per hydrogen atom, for a total of 9 electrons, and 1 mol of hydroxyl contains $9 \mathrm {~N} _ { \mathrm { A } }$ electrons, so A is correct; B. If Fe reacts with sulfur to form FeS, 2 mol of electrons are transferred; if it reacts with $\mathrm { Cl } _ { 2 }$ to form $\mathrm { FeCl } _ { 3 }$, 3 mol of electrons are transferred; the title does not specify the reactants, and the conditions are not sufficient, B is wrong; C. $\mathrm { NH } _ { 3 }$ is dissolved in water and exists partly as $\mathrm { NH } _ { 3 }$, and the total amount of $\mathrm { NH } _ { 3 } \cdot \mathrm { H } _ { 2 } \mathrm { O }$ and $\mathrm { NH } _ { 4 } ^ { + }$ is less than 0.1 mol, which is C wrong; D. The negative reaction in a zinc-copper primary cell is $\mathrm { Zn } - 2 \mathrm { e } ^ { - } = \mathrm { Zn } ^ { 2 + }$, which reduces $65 \mathrm {~g} \mathrm { Zn } ( 1 \mathrm {~mol} )$ and transfers $2 \mathrm {~N} _ { \mathrm { A } }$ electrons, D is wrong;

Question 13: 14. A mixed solution consisting of potassium sulfate, aluminum sulfate, and sulfuric acid with $\mat...

14. A mixed solution consisting of potassium sulfate, aluminum sulfate, and sulfuric acid with $\mathrm { c } \left( \mathrm { H } ^ { + } \right) = 0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } , \mathrm { c } \left( \mathrm { K } ^ { + } \right) = 0.6 \mathrm { mol } \cdot \mathrm { L } ^ { - 1 } , \mathrm { c } \left( \mathrm { SO } _ { 4 } { } ^ { 2 - } \right) = 0.8 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$, then $\mathrm { c } \left( \mathrm { Al } ^ { 3 + } \right)$ is

  • A. A. $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$
  • B. B. $0.3 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$
  • C. C. $0.6 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$
  • D. D. $0.9 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$

Answer: B

Solution: A mixed solution of potassium sulfate, aluminum sulfate, and sulfuric acid consists of $\mathrm { c } \left( \mathrm { H } ^ { + } \right) = 0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } , \mathrm { c } \left( \mathrm { K } ^ { + } \right) = 0.6 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } , \mathrm { c } \left( \mathrm { SO } _ { 4 } { } ^ { 2 - } \right) = 0.8 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ , which is given by conservation of charge, $\mathrm { c } \left( \mathrm { H } ^ { + } \right) + 3 \mathrm { c } \left( \mathrm { Al } ^ { 3 + } \right) + \mathrm { c } \left( \mathrm { K } ^ { + } \right) = \mathrm { c } ( \mathrm { OH } \left. { } ^ { - } \right) + 2 \mathrm { c } \left( \mathrm { SO } _ { 4 } { } ^ { 2 - } \right)$ , and since the concentration of hydroxide is small and negligible, $\mathrm { c } \left( \mathrm { Al } ^ { 3 + } \right) = \left[ 2 \mathrm { c } \left( \mathrm { SO } _ { 4 } { } ^ { 2 - } \right) - \mathrm { c } \left( \mathrm { H } ^ { + } \right) - \right.$ c $\left. \left( \mathrm { K } ^ { + } \right) \right] / 3 = 0.3 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ , and the answer is choice B. .

Question 14: 15. The density of a magnesium chloride solution is $1.18 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ and...

15. The density of a magnesium chloride solution is $1.18 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ and the mass fraction of magnesium ions is $10 \% , 120 \mathrm {~mL}$ The amount of chloride ions in the solution is equal to ( )

  • A. A. 0.58 mol
  • B. B. 1.18 mol
  • C. C. 1.00 mol
  • D. D. 1.50 mol

Answer: B

Solution: The mass of the magnesium chloride solution is: $1.18 \mathrm {~g} \cdot \mathrm {~cm} ^ { - 3 } \times 120 \mathrm {~mL} = 141.6 \mathrm {~g}$ , the amount of magnesium ions is: $( 141.6 \mathrm {~g} \times 10 \% ) / 24 \mathrm {~g} / \mathrm { mol } = 0.59 \mathrm {~mol}$ , and according to conservation of charge, the magnesium chloride solution contains $\mathrm { n } \left( \mathrm { Cl } ^ { - } \right) = 2 \mathrm { n } \left( \mathrm { Mg } ^ { 2 + } \right) = 2 \times 0.59 \mathrm {~mol} =$ 1.18 mol ; so $B$ satisfies the question; So the correct answer to this question : B .

Question 15: 16. Use $\mathrm { N } _ { \mathrm { A } }$ to represent the value of Avogadro's constant. The follo...

16. Use $\mathrm { N } _ { \mathrm { A } }$ to represent the value of Avogadro's constant. The following statement is incorrect ( )

  • A. A. Number of $106 \mathrm { gNa } _ { 2 } \mathrm { CO } _ { 3 }$ in $\mathrm { Na } ^ { + }$ is $2 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. The number of molecules contained in $22.4 \mathrm { LCCl } _ { 4 }$ is $\mathrm { N } _ { \mathrm { A } }$ in standard conditions.
  • C. C. The number of electrons contained in 17 g of ammonia and 17 g of hydroxide are $10 \mathrm {~N} _ { \mathrm { A } }$
  • D. D. Both $24 \mathrm { gO } _ { 2 }$ and $24 \mathrm { gO } _ { 3 }$ contain $1.5 \mathrm {~N} _ { \mathrm { A } }$ of oxygen atoms.

Answer: B

Solution: A. According to $\mathrm { N } = \frac { \mathrm { m } } { \mathrm { M } } \mathrm { N } _ { \mathrm { A } }$, $\mathrm { N } \left( \mathrm { NaCO } _ { 3 } \right) = \mathrm { N } _ { \mathrm { A } }$ can be obtained by bringing in the data, and because of $\mathrm { N } \left( \mathrm { Na } ^ { + } \right) = 2 \mathrm {~N} \left( \mathrm { NaCO } _ { 3 } \right) = 2 \mathrm {~N} _ { \mathrm { A } }$, A is not selected; B. $\mathrm { CCl } _ { 4 }$ is a liquid at standard conditions, so the molar volume of the gas does not apply; C. Ammonia and hydroxide are $10 e ^ { - }$ structures and have molar masses of $17 \mathrm {~g} / \mathrm { mol }$, and the number of electrons in 17 g of ammonia and 17 g of hydroxide ions can be obtained from $\mathrm { N } = \frac { \mathrm { m } } { \mathrm { M } } \mathrm { N } _ { \mathrm { A } }$ by $N \left( \mathrm { NH } _ { 3 } \right) = \mathrm { N } \left( \mathrm { OH } ^ { - } \right) = 10 \mathrm {~N} _ { \mathrm { A } }$, so C is not chosen; D. The number of oxygen atoms in oxygen is two times that of oxygen, and the number of oxygen atoms in ozone is three times that of ozone molecules. [D. The number of oxygen atoms in oxygen is two times that of oxygen, and the number of oxygen atoms in ozone is three times that of ozone molecules, and the number of oxygen atoms in ozone is three times that of ozone molecules, and the number of oxygen atoms in ozone molecules is three times that of ozone molecules; The answer is B.

Question 16: 17. The following statement about NaCl solution of $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$...

17. The following statement about NaCl solution of $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ is correct

  • A. A. Solution containing 1 mol NaCl
  • B. B. 1 L of solution contains 58.5 g NaCl
  • C. C. 1 mol NaCl dissolved in 1 L of water
  • D. D. 58.5 grams of NaCl dissolved in 941.5 grams of water.

Answer: B

Solution: ANALYSIS: $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { NaCl }$ solution means that 1 mol of sodium chloride is dissolved per liter of solution. That is, 1 mol of sodium chloride is dissolved in In other words, 1 mol of sodium chloride dissolved in water makes 1 L of solution with sodium chloride concentration $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$. Analyze the solution according to $\mathrm { c } = \frac { n } { V }$. A. The amount of solute is related to the volume of the solution and the concentration of the solution. The volume of the solution is not told, so it is impossible to calculate the amount of sodium chloride, so A is wrong; B. The amount of substance of 58.5 grams of NaCl is 1 mol, the volume of solution is 1 L, and the concentration of sodium chloride solution is $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$, so option B is correct; C. 1 mol of sodium chloride is dissolved in water to make 1 L of solution, the volume of solution is 1 L, not the volume of solvent, so B is wrong; D. 58.5 grams of NaCl is $1 \mathrm {~mol} , 58.5$ grams of NaCl dissolved in 941.5 g of water in the solution of the mass of 1000 g, the density of the solution is greater than the water, so the volume of the solution is less than 1 L, the concentration is greater than the $1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$, so the D error; So $B$ is correct.

Question 17: 18. The following statements are incorrect ( ) bits) are numerically equal

18. The following statements are incorrect ( ) bits) are numerically equal

  • A. A. The molar mass of phosphoric acid $\mathrm { g } \cdot \mathrm { mol } ^ { - 1 }$ in units of $\left( \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 } \right)$ is related to the mass of $\mathrm { N } _ { \mathrm { A } }$ molecules of phosphoric acid (in g).
  • B. B. The mass ratio of $6.02 \times 10 ^ { 23 }$ nitrogen molecules to $6.02 \times 10 ^ { 23 }$ hydrogen molecules is equal to $14 : 1$
  • C. C. 32 g of oxygen contains approximately $2 \times 6.02 \times 10 ^ { 23 }$
  • D. D. The volume occupied by $0.5 \times 6.02 \times 10 ^ { 23 }$ carbon monoxide molecules at room temperature and pressure is 11.2 L

Answer: D

Solution: A. The molar mass of phosphoric acid is $98 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 }$, and the mass of 1 mol of phosphoric acid is 98 g. The two are numerically equal, but the units are different, so A is correct; B. $6.02 \times 10 ^ { 23 }$ molecules of nitrogen and $6.02 \times 10 ^ { 23 }$ molecules of hydrogen are 1 mol of $\mathrm { N } _ { 2 }$ and $\mathrm { H } _ { 2 }$ respectively, and the mass ratio of the two molecules is 28 $\left. \mathrm { g } \cdot \mathrm { mol } ^ { - 1 } \times 1 \mathrm {~mol} \right) : \left( 2 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 } \times 1 \mathrm {~mol} \right) = 14 : 1$. ], so B is correct; C. The material amount of 32 g of oxygen is $32 \mathrm {~g} / 32 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 } = 1 \mathrm {~mol}$, and each oxygen $\left( \mathrm { O } _ { 2 } \right)$ molecule contains 2 oxygen atoms, so the number of atoms contained in 32 g of oxygen is about $2 \times 6.02 \times 10 ^ { 23 }$, so C is correct; D. Ignoring the fact that the molar volume of the gas is $22.4 \mathrm {~L} \cdot \mathrm {~mol} ^ { - 1 }$, the use of the gas is based on the standard condition, which does not meet the conditions of the question, so D is wrong.

Question 18: 19. The following statements are true for laboratory preparation of $450 \mathrm {~mL} 1 \mathrm {~m...

19. The following statements are true for laboratory preparation of $450 \mathrm {~mL} 1 \mathrm {~mol} / \mathrm { L } \mathrm { NaOH }$ solutions.

  • A. A. Weigh 18 g of NaOH solid on a pallet balance.
  • B. B. Wash the volumetric flasks with distilled water and let them dry before use.
  • C. C. The role of the glass rod in the preparation process is mainly to stir and divert the water
  • D. D. After the volume was fixed and shaken well, it was found that the liquid level dropped, continue to add water to the scale line

Answer: C

Solution: A. A 500 mL volumetric flask is used to prepare $450 \mathrm {~mL} 1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { NaOH }$ solution in the laboratory. The mass of sodium hydroxide solid $m = M \cdot n = c v M = 0.5 \mathrm {~L} \times 1 \mathrm {~mol} / \mathrm { L } \times 40 \mathrm {~g} / \mathrm { mol } = 20.0 \mathrm {~g}$ should be weighed on an electronic balance to get 20.0 g of NaOH solid, so A is wrong; B. volumetric flasks can be used directly after washing with distilled water, do not need to dry, so B error; C. glass rod in the dissolution of solids stirring can accelerate the dissolution of solids, transfer of liquids to play the role of drainage, so C is correct; D. After shaking the volume, found that the liquid level drops, continue to add water to the scale line, will lead to the solution volume is large, the solution concentration is low, so D error;

Question 19: 20. Given ${ } ^ { \mathrm { N } _ { \mathrm { A } } }$ as the value of Avogadro's constant, one of ...

20. Given ${ } ^ { \mathrm { N } _ { \mathrm { A } } }$ as the value of Avogadro's constant, one of the following statements is correct

  • A. A. The number of polar covalent bonds in $0.1 \mathrm {~mol} \mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ at room temperature and pressure is $0.1 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. $100 \mathrm { ml } 18.4 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ Sulfuric acid reacts with sufficient copper to produce ${ } ^ { \mathrm { SO } _ { 2 } }$ in $0.92 \mathrm {~N} _ { \mathrm { A } }$ molecules
  • C. C. Equal volume, equal concentration of $\mathrm { NaCl } _ { \text {和 } } \mathrm { KCl } _ { \text {溶液中,阴,阳离子数目之和均约为 } } 2 \mathrm {~N} _ { \mathrm { A } }$
  • D. D. $1.2 \mathrm {~g} \mathrm { NaHSO } _ { 4 }$ The total number of cations and anions in the crystal is $0.02 \mathrm {~N} _ { \mathrm { A } }$

Answer: D

Solution: A. The $0.1 \mathrm { molH } _ { 2 } \mathrm { O } _ { 2 }$ molecule contains $0.2 \mathrm { molO } - \mathrm { H }$ polar covalent bonds, and the number of polar covalent bonds is 0.2NA, so A is wrong; B. The amount of $100 \mathrm { ml } 18.4 \mathrm {~mol} / \mathrm { LH } _ { 2 } \mathrm { SO } _ { 4 }$ is 1.84 mol, and according to $\mathrm { Cu } + 2 \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } ($, $) \stackrel { ( } { = } = \mathrm { CuSO } _ { 4 } + \mathrm { SO } _ { 2 }$ is concentrated. According to $\mathrm { Cu } + 2 \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } ($, the amount of $) \stackrel { ( } { = } = \mathrm { CuSO } _ { 4 } + \mathrm { SO } _ { 2 }$ is 1.84 mol, and if 1.84 mol of sulfuric acid reacts completely, the amount of $\mathrm { SO } _ { 2 }$ will be 0.92 mol. However, as the reaction proceeds, the concentration of sulfuric acid decreases, and the dilute sulfuric acid and the copper don't react, so the amount of $\mathrm { SO } _ { 2 }$ will be less than 0.92 mol; C. In equal volumes and equal concentrations of " NaCl" and " KCl" solutions, the number of anions and cations cannot be calculated because the concentrations and volumes of the solutions are unknown, so C is wrong; D.1.2 $\mathrm { gNaHSO } _ { 4 }$ is $0.01 \mathrm {~mol} , \mathrm { NaHSO } _ { 4 }$ The amount of substance in the crystal is $0.01 \mathrm {~mol} , \mathrm { NaHSO } _ { 4 }$ There are sodium ions and bisulfate ions in the crystal, and the total number of cations and anions is, so D is correct; The answer to this question is D. [[EYE POINT]] Copper and concentrated sulfuric acid react when heated, a certain amount of concentrated sulfuric acid, as the reaction proceeds, the concentration decreases, when changed to dilute sulfuric acid after the reaction is stopped, so the amount of sulfur dioxide generated, can not be calculated by the amount of the original sulfuric acid, should be participating in the reaction of sulfuric acid (sulfuric acid showing oxidizing properties) shall prevail.

Question 20: 21. The following is a correct understanding of the word "mole".

21. The following is a correct understanding of the word "mole".

  • A. A. A mole is a unit of mass.
  • B. B. A mole is a unit of quantity of a substance, symbolized as mol
  • C. C. The mole is a fundamental physical quantity
  • D. D. The mole is the unit of mass of a substance

Answer: B

Solution: A. The mole is the unit of quantity of a substance, A is wrong ; B. The unit of quantity of a substance is the mole, abbreviated as moles, and symbolized as mol, B correct; C. The quantity of matter is one of the seven physical quantities of the International System of Units, and its unit is the mole, C is wrong; D. The mole is the unit of the amount of matter, and the unit of mass is the gram, D is wrong; The answer is B. [Eyes on] this question examines the amount of matter and its unit, clear the concept of the amount of matter and the unit is the key to solving the problem, pay attention to the amount of matter to use notes, the topic is not difficult.

Question 21: 22. Nitrogen oxides and carbon monoxide in the exhaust of automobiles cause air pollution, and catal...

22. Nitrogen oxides and carbon monoxide in the exhaust of automobiles cause air pollution, and catalytic purification devices are usually used to remove them. Catalyst Catalytic purification devices are usually used to convert them into non-toxic substances and then emit them. The main reaction principle is $2 \mathrm { NO } + 2 \mathrm { CO } \quad \mathrm { X } + 2 \mathrm { CO } _ { 2 }$, please guess the chemical formula of X.

  • A. A. $\mathrm { O } _ { 2 }$
  • B. B. $\mathrm { N } _ { 2 }$
  • C. C. $\mathrm { H } _ { 2 } \mathrm { O }$
  • D. D. $\mathrm { NO } _ { 2 }$

Answer: B

Solution: According to the law of conservation of mass, before and after the chemical reaction, the type and number of atoms remain the same and the reactants contain N , C, the number of $O$ is 2, 2, 4 respectively, and the number of $C$, $O$ in the product $C _ { 2 }$ is 2, 4 respectively, so $X$ should also contain 2 $N$ atoms. ] should contain 2 $N$ atoms, so the chemical formula of $X$ is $N _ { 2 }$. Therefore, $B$ is chosen.

Question 22: 23. A small amount of sodium metal is added to anhydrous ethanol and a reaction occurs $2 \mathrm { ...

23. A small amount of sodium metal is added to anhydrous ethanol and a reaction occurs $2 \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { OH } + 2 \mathrm { Na } \rightarrow 2 \mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { ONa } + \mathrm { H } _ { 2 } \uparrow \quad , N _ { \mathrm { A } }$ is the value of Avogadro's constant, the following statement is correct

  • A. A. The total number of electrons in $0.1 \mathrm {~mol} - \mathrm { CH } _ { 2 } { } ^ { - }$ is $0.8 N _ { \mathrm { A } }$
  • B. B. $4.6 \mathrm { gCH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { OH }$ The number of oxygen molecules consumed in complete combustion is $0.35 N _ { \mathrm { A } }$
  • C. C. In $0.1 \mathrm { molCH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { ONa }$, the number of polar keys is $0.7 N _ { \mathrm { A } }$
  • D. D. Number of electrons transferred by the reaction per 11.2 L of hydrogen produced under $25 ^ { \circ } \mathrm { C } , 101 \mathrm { kPa }$ $N _ { \mathrm { A } }$

Answer: A

Solution: A. Each ${ } ^ { - \mathrm { CH } _ { 2 } - }$ group contains 1 carbon atom (6 electrons) and 2 hydrogen atoms (1 electron each), and the total number of electrons in $6 + 2 \times 1 = 8,0.1 \mathrm {~mol} ^ { - \mathrm { CH } _ { 2 } - }$ is $0.1 \times 8 \mathrm {~N} _ { \mathrm { A } } = 0.8 \mathrm {~N} _ { \mathrm { A } }$, which is correct; B. The chemical equation for the complete combustion of 4.6 g of ethanol (0.1 mol) is $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { OH } + 3 \mathrm { O } _ { 2 } \xrightarrow { \text { ignite } } 2 \mathrm { CO } _ { 2 } + 3 \mathrm { H } _ { 2 } \mathrm { O }$, which consumes $0.3 \mathrm {~mol} { } ^ { \mathrm { O } _ { 2 } }$ and corresponds to $0.3 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ of ${ } ^ { \mathrm { O } _ { 2 } }$ molecules, not $0.35 { } ^ { \mathrm { N } _ { \mathrm { A } } }$. FORMULA_7]], B is wrong; C. The number of polar bonds in $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { ONa }$ includes one $\mathrm { C } - \mathrm { O }$ bond and five $\mathrm { C } - \mathrm { H }$ bonds (six in total), and the number of polar bonds in $0.1 \mathrm { molCH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { ONa }$ is $0.1 \times 6 \mathrm {~N} _ { \mathrm { A } } = 0.6 \mathrm {~N} _ { \mathrm { A } }$, rather than $0.1 \times 6 \mathrm {~N} _ { \mathrm { A } } = 0.6 \mathrm {~N} _ { \mathrm { A } }$. ], not $0.7 N _ { \mathrm { A } }$, and C is incorrect; D. Under $25 ^ { \circ } \mathrm { C } , 101 \mathrm { kPa }$, the molar volume of gas is greater than $22.4 \mathrm {~L} / \mathrm { mol } , 11.2 \mathrm {~L} ^ { \mathrm { H } _ { 2 } }$ in an amount less than 0.5 mol, and 2 mol of electrons are transferred for every $1 \mathrm {~mol} { } ^ { \mathrm { H } _ { 2 } }$, so the number of electrons transferred is less than ${ } ^ { N _ { \mathrm { A } } }$, D is wrong. ], D is wrong.

Question 23: 24. It is known that the smaller the molar mass of a gas, the faster it diffuses. The figure below s...

24. It is known that the smaller the molar mass of a gas, the faster it diffuses. The figure below shows an experiment on the diffusion rate of gases, two gas Diffusion of two gases meet to form a white smoke ring. The following about the substance a, b judgment is correct ( ) ![](/images/questions/mole-calculation/image-001.jpg)

  • A. A. A is concentrated ammonia, B is concentrated sulfuric acid
  • B. B. A is concentrated hydrochloric acid and B is concentrated ammonia.
  • C. C. A is concentrated ammonia, B is concentrated hydrochloric acid
  • D. D. A is concentrated sulfuric acid and B is concentrated ammonia.

Answer: C

Solution: Concentrated sulfuric acid does not volatilize, eliminating the AD option. Since the smaller the molar mass of a gas, the faster it diffuses, ammonia diffuses faster than hydrogen chloride, and ammonia is farther away from the smoke ring than concentrated hydrochloric acid, so A is concentrated ammonia and B is concentrated hydrochloric acid.

Question 24: 25. A NaCl solution of $500 \mathrm {~mL} 0.40 \mathrm {~mol} / \mathrm { L }$ is prepared in the la...

25. A NaCl solution of $500 \mathrm {~mL} 0.40 \mathrm {~mol} / \mathrm { L }$ is prepared in the laboratory from NaCl solid. The following statements are true

  • A. A. Weigh 23.4 g of NaCl solid.
  • B. B. The NaCl solid was placed in a 500 mL volumetric flask and dissolved with water.
  • C. C. The glass rod acts as an accelerator of dissolution and drainage
  • D. D. Looking down at the graduated line of the volumetric flask during the finalization of the volume will result in a low concentration of the prepared solution

Answer: C

Solution: A. To prepare a NaCl solution of $500 \mathrm {~mL} 0.40 \mathrm {~mol} / \mathrm { L }$, the mass of NaCl used is: $58.5 \mathrm {~g} / \mathrm { mol } \times 0.5 \mathrm {~L} \times 0.40 \mathrm {~mol} / \mathrm { L } = 11.7 \mathrm {~g}$, so A is incorrect; B. The volumetric flask can not be used to dissolve solids, should be dissolved in the beaker, and then pipetted to the volumetric flask, so B error; C. In the process of dissolution, the glass rod plays the role of stirring to accelerate the role of dissolution, in the transfer of the solution, the glass rod plays the role of drainage, so C is correct; D. When fixing the volume, looking down on the scale, the liquid surface below the scale, the volume of the solution is small, the concentration of the solution is high, so D error;

Question 25: 26. Let ${ } ^ { N _ { A } }$ be the value of Avogadro's constant. The following statement is correc...

26. Let ${ } ^ { N _ { A } }$ be the value of Avogadro's constant. The following statement is correct

  • A. A. $16.9 \mathrm {~g} _ { \text {过氧化钡 } } \left( \mathrm { BaO } _ { 2 } \right)$ The total number of anions and cations in the solid is $0.2 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. The mass of $\mathrm { N } _ { \mathrm { A } }$ iron hydroxide colloidal particles is ${ } ^ { 107 \mathrm {~g} }$
  • C. C. Hydrocarbon molecules of $70 \mathrm { gC } _ { 5 } \mathrm { H } _ { 10 }$ contain $\mathrm { C } - \mathrm { C } _ { \text {共价键的数目一定为 } } 4 \mathrm {~N} _ { \mathrm { A } }$
  • D. D. At room temperature, the number of hydrogen ions contained in an acetic acid solution of $0.1 \mathrm {~mol} / \mathrm { L }$ is $0.1 \mathrm {~N} _ { \mathrm { A } }$

Answer: A

Solution: A. The amount of barium peroxide in 16.9 g is 0.1 mol, and the total number of anions and cations in the solid is 0.2 mol, totaling $0.2 \mathrm {~N} _ { \mathrm { A } }$, so A is correct; B. A ferric hydroxide pellet is an aggregate of several ferric hydroxides, then $\mathrm { N } _ { \mathrm { A } }$ a ferric hydroxide pellet contains more than 1 mol of ferric hydroxide, and the mass is more than 107 g, so B is wrong; C. It is not certain whether the substance is a cycloalkane or an unsaturated hydrocarbon, so the number of C-C covalent bonds cannot be calculated, so C is wrong; D. The volume of the acetic acid solution is not given, so the number of hydrogen ions cannot be calculated, so D is wrong;

Question 26: 27. Scientists have produced a new molecule ${ } ^ { \mathrm { O } _ { 4 } }$, and the following sta...

27. Scientists have produced a new molecule ${ } ^ { \mathrm { O } _ { 4 } }$, and the following statements about ${ } ^ { \mathrm { O } _ { 4 } }$ and ${ } ^ { \mathrm { O } _ { 2 } }$ are incorrect

  • A. A. stereoisomorphic
  • B. B. Equal masses of ${ } ^ { \mathrm { O } _ { 4 } }$ and ${ } ^ { \mathrm { O } _ { 2 } }$ contain different numbers of atoms.
  • C. C. All are monomers
  • D. D. Their transformation is a chemical change

Answer: B

Solution: A. Monomers of the same element are called isomers of each other. ${ } ^ { \mathrm { O } _ { 4 } }$ and ${ } ^ { \mathrm { O } _ { 2 } }$ are both monomers of the element oxygen, and are isomers; B. $\mathrm { O } _ { 4 }$ and $\mathrm { O } _ { 2 }$ are monomers of the element oxygen, and equal masses of $\mathrm { O } _ { 4 }$ and $\mathrm { O } _ { 2 }$ contain the same number of oxygen atoms, e.g. 16 g of $\mathrm { O } _ { 4 }$. FORMULA_6]] and $\mathrm { O } _ { 2 }$ both contain the same number of oxygen atoms; C. $\mathrm { O } _ { 4 }$ and $\mathrm { O } _ { 2 }$ are both monomers of the element oxygen; D. $\mathrm { O } _ { 4 }$ and $\mathrm { O } _ { 2 }$ are two different kinds of monomers, and their transformation is a chemical change in which new substances are formed. The answer is B.

Question 27: 28. $N _ { \mathrm { A } }$ is the value of Avogadro's constant, and the following statement is corr...

28. $N _ { \mathrm { A } }$ is the value of Avogadro's constant, and the following statement is correct

  • A. A. The number of $\mathrm { H } ^ { + }$ in 22.4 L of HCl gas under standard conditions is $N _ { \mathrm { A } }$
  • B. B. 2.4 g graphite crystals contain the number of C-C bonds $0.6 N _ { \mathrm { A } }$
  • C. C. 4.6 g Na reacts fully with $0.5 \mathrm {~L} 0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ dilute sulfuric acid and the number of electrons transferred is $0.2 N _ { \mathrm { A } }$
  • D. D. If $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ is known to be a positive salt, then $1 \mathrm {~mol} \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ contains the hydroxyl group number $2 N _ { \mathrm { A } }$

Answer: C

Solution: $\mathrm { A } . \mathrm { HCl }$ The gas is made up of molecules and is not ionized, no $\mathrm { H } ^ { + } , \mathrm { A }$ error; B. The amount of 2.4 g of graphite is 0.2 mol. According to the structure of graphite, 1 mol of C contains 1.5 mol of covalent bonds, so 0.2 mol of graphite contains $0.3 N _ { A }$ of C-C bonds, and B is wrong; C. 4.6 g of Na has a substance amount of 0.2 mol and reacts with dilute sulfuric acid. Even if the amount of dilute sulfuric acid is not enough, the water reacts with the sodium, so the Na reacts completely to form $\mathrm { H } _ { 2 } 0.1 \mathrm {~mol}$, and the number of electrons transferred is $0.2 N _ { A }$, and C is correct; D. Hydroxyl hydrogen atoms in oxygen-containing acids are acidic, $\mathrm { NaH } _ { 2 } \mathrm { PO } _ { 2 }$ is an ortho-salt, $\mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ is a monoacid, so $1 \mathrm {~mol} \mathrm { H } _ { 3 } \mathrm { PO } _ { 2 }$ contains $N _ { A }$ of hydroxyl groups, D is wrong; The question is C.

Question 28: 29. The following experimental setup, reagent selection or operation is correct.

29. The following experimental setup, reagent selection or operation is correct.

  • A. A. ![](/images/questions/mole-calculation/image-001.jpg)
  • B. B. ![](/images/questions/mole-calculation/image-002.jpg)
  • C. C. ![](/images/questions/mole-calculation/image-003.jpg)
  • D. D. ![](/images/questions/mole-calculation/image-004.jpg)

Answer: D

Solution: A. Since $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } + \mathrm { CO } _ { 2 } = 2 \mathrm { NaHCO } _ { 3 }$, saturated sodium carbonate solution cannot be used to remove HCl from $\mathrm { CO } _ { 2 }$, saturated sodium bicarbonate solution should be used, A is wrong; B. The volumetric flask is an instrument that cannot be used for dissolving, diluting, or reacting; C. The laboratory formula for ammonia is $2 \mathrm { NH } _ { 4 } \mathrm { Cl } + \mathrm { Ca } ( \mathrm { OH } ) _ { 2 } \quad \mathrm { CaCl } _ { 2 } + 2 \mathrm { NH } _ { 3 } \uparrow + 2 \mathrm { H } _ { 2 } \mathrm { O }$, C is wrong; D. Chlorine does not react with concentrated sulfuric acid, and concentrated sulfuric acid has a water-absorbent can be used as a desiccant, D is correct; Therefore, the answer is D.

Question 29: 30. Astronaut Wang Yaping, while teaching in space, touched a weightless ball of supersaturated sodi...

30. Astronaut Wang Yaping, while teaching in space, touched a weightless ball of supersaturated sodium acetate solution with a cotton swab with a crystalline nucleus stuck to it, causing it to rapidly crystallize into a glowing ball of crystals. The following statement is correct. ![](/images/questions/mole-calculation/image-002.jpg)

  • A. A. Sodium acetate can only be made by reacting acetic acid with sodium hydroxide
  • B. B. Sodium acetate and sodium chloride solutions have the same ability to conduct electricity
  • C. C. Sodium acetate is both a sodium and acetate salt
  • D. D. Desired Formulation ${ } ^ { 1.00 \mathrm {~L} 1.00 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \text { 的 } \mathrm { CH } _ { 3 } \mathrm { COONa } \text { 溶液,可将 } { } ^ { 8.2 \mathrm { gCH } _ { 3 } \mathrm { COONa } } \text { 溶于 } { } ^ { 1.00 \mathrm {~L} } \text { 水中 } }$

Answer: C

Solution: A. Sodium acetate can also be produced by the reaction of sodium oxide and acetic acid and the reaction of sodium carbonate and acetic acid; B. The strength of the conductivity of the solution is related to factors such as the size of the concentration of free-moving ions, and the concentration of the solution is unknown, so it is impossible to compare, B error; C. Sodium acetate belongs to both sodium salt and acetate, C is correct; D. To prepare ${ } ^ { 1.00 \mathrm {~L} 1.00 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \text { 的 } \mathrm { CH } _ { 3 } \mathrm { COONa } \text { 溶液,可将 } { } ^ { 8.2 \mathrm { gCH } _ { 3 } \mathrm { COONa } } \text { 溶于水配成 } 1 \mathrm {~L} \text { 溶液,} }$ 1 L of water is not equal to the volume of 1 L of solution, D error; Therefore, the answer is C.

Question 30: 31. Let ${ } ^ { N _ { \mathrm { A } } }$ denote the value of Avogadro's constant. (1) The number of...

31. Let ${ } ^ { N _ { \mathrm { A } } }$ denote the value of Avogadro's constant. (1) The number of carbon atoms in $11.2 \mathrm {~L} ^ { \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH } }$ is $N _ { \mathrm { A } }$ in standard conditions. (2) 6.9 g of Na reacts completely with ${ } ^ { \mathrm { O } _ { 2 } }$, the number of electrons transferred is $0.3 N _ { \mathrm { A } }$. (3) The total number of electrons transferred when $11.2 \mathrm {~L} ^ { \mathrm { Cl } _ { 2 } }$ is dissolved in water is $0.5 N _ { \mathrm { A } }$ under standard conditions. (4) The number of protons in 18g of $\mathrm { D } _ { 2 } \mathrm { O }$ and 18g of $\mathrm { H } _ { 2 } \mathrm { O }$ is $10 N _ { \mathrm { A } }$. (5) The total number of oxygen atoms in $2 l g$ oxygen and $27 g$ ozone at room temperature and pressure is $3 N _ { A }$. (6) $40 \mathrm {~mL} 10 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ Concentrated hydrochloric acid reacts sufficiently with a sufficient amount of $\mathrm { MnO } _ { 2 }$ to form $\mathrm { Cl } _ { 2 }$, the number of molecules is $0.1 N _ { \mathrm { A } }$.

  • A. A. 1
  • B. B. 2
  • C. C. 3
  • D. D. 4

Answer: B

Solution: (1) standard condition, ethanol is not gaseous, not to determine the amount of its substance, error; (2) 6.9 g of Na is 0.3 mol, and the sodium valence changes from 0 to + 1 in the reaction, so the complete reaction with ${ } ^ { \mathrm { O } _ { 2 } }$ transfers 0.3 mol of electrons, and the number of electrons is ${ } ^ { 0.3 N _ { \mathrm { A } } }$, correct; (3) Under standard conditions, 11.2 L of $\mathrm { Cl } _ { 2 }$ (0.5 mol) is dissolved in water, and the total number of electrons transferred is less than $0.5 { } ^ { N _ { \mathrm { A } } }$ because chlorine gas partially reacts with water, which is incorrect; (4) The amounts of protons contained in $18 \mathrm { gD } _ { 2 } \mathrm { O }$ (0.9 mol) and $18 \mathrm { gH } _ { 2 } \mathrm { O }$ (1 mol) are 9 mol and 10 mol, respectively, which is wrong; (5) oxygen and ozone are oxygen atoms composed of a single substance, at room temperature and pressure, 21 g of oxygen and 27 g of ozone contains the amount of oxygen atoms $48 \mathrm {~g} \div 16 \mathrm {~g} / \mathrm { mol } = 3 \mathrm {~mol}$, the total number of oxygen atoms is $3 ^ { N _ { \mathrm { A } } }$, correct; (6) As the reaction proceeds, the concentration of hydrochloric acid decreases and the reaction no longer proceeds, so the chlorine gas generated is less than 0.1 mol, which is incorrect; the answer is B.

Question 31: 33. The production of sodium oxide from sodium metal usually utilizes the following reactions: $2 \m...

33. The production of sodium oxide from sodium metal usually utilizes the following reactions: $2 \mathrm { NaNO } _ { 3 } + 10 \mathrm { Na } = 6 \mathrm { Na } _ { 2 } \mathrm { O } + \mathrm { N } _ { 2 } \uparrow , 1 \mathrm { molNa }$ with. $0.3 \mathrm { molNaNO } _ { 3 }$ When fully reacted, the following statement is false $0.1 \mathrm { molNaNO } _ { 3 }$

  • A. A. There must be ${ } ^ { 0.6 \mathrm { molNa } _ { 2 } \mathrm { O } }$ and ${ } ^ { 2.24 \mathrm { LN } _ { 2 } }$ generated.
  • B. B. reactive residue
  • C. C. Transfer of electrons ${ } ^ { N _ { \mathrm { A } } }$ during the reaction
  • D. D. The product contains 0.6 mol of anions

Answer: A

Solution: A. The temperature and pressure are not specified, so the volume of nitrogen generated cannot be calculated, so A is wrong; B. According to the reaction equation, 1 molNa consumes $0.2 \mathrm { molNaNO } _ { 3 }$, so there is $0.1 \mathrm { molNaNO } _ { 3 }$ left in the reaction, so B is correct; C. In this reaction, the valence of sodium increases from 0 to +1, consuming 1 molNa and transferring 1 mol of electrons, so $N _ { A }$ electrons are transferred, so C is correct; D. $\mathrm { Na } _ { 2 } \mathrm { O }$ is composed of sodium ions and oxygen ions, and there are no ions in nitrogen. D. $\mathrm { Na } _ { 2 } \mathrm { O }$ is made up of sodium ions and oxygen ions, and there are no ions in nitrogen. 1 molNa takes part in the reaction to form $0.6 \mathrm { molNa } _ { 2 } \mathrm { O } , ~ 1 \mathrm { molNa } _ { 2 } \mathrm { O }$, which contains $1 \mathrm { molO } ^ { 2 - }$, and so $0.6 \mathrm { molNa } _ { 2 } \mathrm { O }$ contains 0.6 mol of the anion, so D is correct; therefore, A is the right choice.

Question 32: 34. Let $N _ { \mathrm { A } }$ be the value of Avogadro's constant. The following statement is corr...

34. Let $N _ { \mathrm { A } }$ be the value of Avogadro's constant. The following statement is correct

  • A. A. The number of oxygen atoms contained in $11.2 \mathrm { LNO } _ { 2 }$ is $N _ { A }$ in standard conditions.
  • B. B. The number of neutrons contained in a mixture of $1 \mathrm {~mol} ^ { 20 } \mathrm { Ne }$ and ${ } ^ { 22 } \mathrm { Ne }$ is $10 N _ { A }$
  • C. C. The number of anions contained in a mixture of $8.4 \mathrm { gNaHCO } _ { 3 }$ and $\mathrm { MgCO } _ { 3 }$ is $0.1 N _ { A }$
  • D. D. If the $\mathrm { pH } = 4.6$ of a saturated solution of boric acid ($\mathrm { H } _ { 3 } \mathrm { BO } _ { 3 }$) at a certain temperature is known, the number of $\mathrm { H } ^ { + }$ in the solution is $1 \times 10 ^ { - 4.6 } N _ { A }$

Answer: C

Solution: A. Under standard conditions, $\mathrm { NO } _ { 2 }$ liquefies and is partially converted to $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$, making it impossible to calculate the number of oxygen atoms contained; A is incorrect; B. A mixture of $1 \mathrm {~mol} ^ { 20 } \mathrm { Ne }$ and ${ } ^ { 22 } \mathrm { Ne }$ contains a number of neutrons between $10 N _ { \mathrm { A } } \sim 12 N _ { \mathrm { A } }$, B is incorrect; C. Both $8.4 \mathrm { gNaHCO } _ { 3 }$ and $8.4 \mathrm {~g} \mathrm { MgCO } _ { 3 }$ contain $0.1 N _ { A }$, so the number of anions in 8.4 g of the mixture should also be $0.1 N _ { A }$, and C is correct; D. The volume of the saturated solution of boric acid is unknown, so the number of $\mathrm { H } ^ { + }$ in the solution cannot be calculated, and D is incorrect. $35 . \mathrm { C }$ [Knowledge Points] ethyl acetate preparation experiment device and operation, the preparation of colloids, the preparation of the experiment and the steps and operations B. Preparation of ${ } ^ { \mathrm { Fe } ( \mathrm { OH } ) _ { 3 } }$ colloid, need to saturated ferric chloride solution drops into boiling water, the title does not have an alcohol lamp, B error; C. Preparation of $100 \mathrm {~mL} 1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ dilute sulfuric acid, the need for experimental equipment: beakers, glass rods, measuring cylinders, 100 mL volumetric flasks, colloid burette, etc., can be completed, C is correct; D. $\mathrm { CuSO } _ { 4 }$ Concentration and crystallization of solutions requires the following experimental apparatus: alcohol lamp, evaporating dish, glass rod, etc. The experiment cannot be completed because of the lack of an evaporating dish and alcohol lamp, D is wrong;

Question 33: 35. The experiment can be accomplished with the following apparatus (clamping device omitted). ![](...

35. The experiment can be accomplished with the following apparatus (clamping device omitted). ![](/images/questions/mole-calculation/image-003.jpg) (1) ![](/images/questions/mole-calculation/image-004.jpg) (2) ![](/images/questions/mole-calculation/image-005.jpg) (3) ![](/images/questions/mole-calculation/image-006.jpg) (4) ![](/images/questions/mole-calculation/image-007.jpg) (5) ![](/images/questions/mole-calculation/image-008.jpg) (6)

  • A. A. Preparation of ethyl acetate
  • B. B. Preparation of $\mathrm { Fe } ( \mathrm { OH } ) _ { 3 }$ Colloids
  • C. C. Preparation of ${ } ^ { 100 } \mathrm {~mL} 1.0 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ dilute sulfuric acid
  • D. D. $\mathrm { CuSO } _ { 4 }$ Concentrated crystallization of solutions

Answer: C

Solution:

Question 34: 36. $\mathrm { N } _ { \mathrm { A } }$ is the value of Avogadro's constant. The following statement...

36. $\mathrm { N } _ { \mathrm { A } }$ is the value of Avogadro's constant. The following statement is correct.

  • A. A. $1 \mathrm { molNa } _ { 2 } \mathrm { O } _ { 2 }$ The total number of ions contained in the solid is $4 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. The volume of $2 \mathrm {~N} _ { \mathrm { A } }$ and $\mathrm { SO } _ { 3 }$ is 44.8 L in standard condition.
  • C. C. At room temperature, the number of atoms contained in a gas mixture consisting of $4.8 \mathrm { gO } _ { 2 }$ and $\mathrm { O } _ { 3 }$ is $0.3 \mathrm {~N} _ { \mathrm { A } }$.
  • D. D. $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ The number of $\mathrm { Na } ^ { + }$ in the solution is $0.2 \mathrm {~N} _ { \mathrm { A } }$

Answer: C

Solution: A. Sodium peroxide is composed of sodium ions and peroxide ions, $1 \mathrm { molNa } _ { 2 } \mathrm { O } _ { 2 }$ solid contains $3 \mathrm {~N} _ { \mathrm { A } }$ total number of ions, A is wrong; B. $\mathrm { SO } _ { 3 }$ is a solid in standard condition, and the volume of $2 \mathrm {~N} _ { \mathrm { A } }$ is not 44.8 L. B is wrong; C. At room temperature, the gas mixture of $4.8 \mathrm { gO } _ { 2 }$ and $\mathrm { O } _ { 3 }$ contains $4.8 \mathrm {~g} \div 16 \mathrm {~g} / \mathrm { mol } = 0.3 \mathrm {~mol}$ atoms in the amount of $4.8 \mathrm {~g} \div 16 \mathrm {~g} / \mathrm { mol } = 0.3 \mathrm {~mol}$, and $0.3 \mathrm {~N} _ { \mathrm { A } }$ in the number of $0.3 \mathrm {~N} _ { \mathrm { A } }$, C is correct; D. The volume of the solution is unknown, and the number of $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ in the $\mathrm { Na } ^ { + }$ solution is not necessarily $0.2 \mathrm {~N} _ { \mathrm { A } }$, and D is wrong; the answer is C.

Question 35: 37. The description of the $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \...

37. The description of the $0.1 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 } \mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ solution is incorrect ().

  • A. A. 0.5 L of this solution, the concentration of $\mathrm { Na } ^ { + }$ is $0.2 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$
  • B. B. 1 L of this solution contains $\mathrm { CO } _ { 3 } { } ^ { 2 - }$ less than $0.1 \mathrm {~N} _ { \mathrm { A } } \left( \mathrm { N } _ { \mathrm { A } } \right.$ is Avogadro's constant $)$
  • C. C. If 100 mL of the solution is removed from 1 L of the solution, the concentration of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ in the removed solution is $0.01 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$.
  • D. D. Take 10 mL of this solution and dilute it to 100 mL with water, the concentration of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ is $0.01 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$.

Answer: C

Solution: A. The material concentration of sodium ions is $0.1 \mathrm {~mol} / \mathrm { L } \times 2 = 0.2 \mathrm {~mol} / \mathrm { L }$, so A is correct; B. There is hydrolysis of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ by $\mathrm { CO } _ { 3 } { } ^ { 2 - }$ in the $\mathrm { CO } _ { 3 } { } ^ { 2 - }$ solution, so the number of $\mathrm { CO } _ { 3 } { } ^ { 2 - }$ is less than the number of $0.1 \mathrm {~N} _ { \mathrm { A } }$ in 1 L of the solution, so B is correct; C. If 100 mL of the solution is removed from 1 L of the solution and the solution is homogeneous, the concentration of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ in the removed solution is $0.1 \mathrm {~mol} / \mathrm { L }$, so C is wrong; D. Take 10 mL of the solution and dilute it to 100 mL with water, the concentration of $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ is $\frac { 0.01 \mathrm {~L} \times 0.1 \mathrm {~mol} / \mathrm { L } } { 0.1 \mathrm {~L} } = 0.01 \mathrm {~mol} / \mathrm { L }$, so D is correct; Therefore, the answer is C.

Question 36: 38. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant. The following state...

38. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant. The following statement is correct

  • A. A. The number of electrons transferred in the complete reaction of 1 molFe with hot water vapor is $3 \mathrm {~N} _ { \mathrm { A } }$
  • B. B. The hydrocarbon molecule of $28 \mathrm {~g} \mathrm { C } _ { 5 } \mathrm { H } _ { 10 }$ contains the number of carbon-carbon double bonds $0.4 \mathrm {~N} _ { \mathrm { A } }$
  • C. C. The number of hydrogen atoms in 7.0 g of a mixture of butene and propylene at room temperature and pressure is $\mathrm { N } _ { \mathrm { A } }$
  • D. D. The $5 \mathrm {~mL} 0.1 \mathrm {~mol} / \mathrm { LCH } _ { 3 } \mathrm { COOH }$ solution and the $10 \mathrm {~mL} 0.5 \mathrm {~mol} / \mathrm { L } \mathrm { CH } _ { 3 } \mathrm { COOH }$ solution contain an equal number of $\mathrm { CH } _ { 3 } \mathrm { COOH }$ molecules.

Answer: C

Solution: A. Fe reacts completely with high-temperature water vapor to form triiron tetraoxide, and the number of electrons transferred from 1 molFe is $\frac { 8 } { 3 } N _ { A }$, which is wrong; B. The hydrocarbon molecule of $\mathrm { C } _ { 5 } \mathrm { H } _ { 10 }$ may be a cycloalkane that does not contain a carbon-carbon double bond, B is wrong; C. Butene and propylene's simplest formula is $\mathrm { CH } _ { 2 }$, at room temperature and pressure, 7.0 g of butene and propylene gas mixture contains 0.5 mol of $\mathrm { CH } _ { 2 }$, then the number of hydrogen atoms contained in the $N _ { \mathrm { A } }$, C narrative is correct; D. Acetic acid is a weak electrolyte, the greater the concentration, the less ionization, $5 \mathrm {~mL} 0.1 \mathrm {~mol} / \mathrm { LCH } _ { 3 } \mathrm { COOH }$ solution and 10 mL of $0.5 \mathrm {~mol} / \mathrm { L } \mathrm { CH } _ { 3 } \mathrm { COOH }$ solution contains $\mathrm { CH } _ { 3 } \mathrm { COOH }$ molecules are not equal, D narrative error; The answer is C.

Question 37: 39. Given $N _ { A }$ as the value of Avogadro's constant, the following statement is correct

39. Given $N _ { A }$ as the value of Avogadro's constant, the following statement is correct

  • A. A. At room temperature and pressure, 1 mol of Cl 2 is dissolved in $1.0 \mathrm { LH } _ { 2 } \mathrm { O }$, and the sum of the numbers of $\mathrm { Cl } ^ { - } , \mathrm { HClO } , \mathrm { ClO }$ in the solution is $2 N _ { A }$.
  • B. B. In the standard condition, $11.2 \mathrm { LH } _ { 2 } \mathrm { O } _ { 2 }$ contains the number of nonpolar covalent bonds as $0.5 N _ { A }$
  • C. C. 56.0 g Fe reacts with $1.0 \mathrm {~mol} \mathrm { Br } _ { 2 }$ and the number of electrons transferred is $2 N _ { A }$
  • D. D. In standard conditions, $11.2 \mathrm { LNH } _ { 3 }$ and 11.2 L HF both contain $5 N _ { A }$ protons

Answer: C

Solution: $\mathrm { A } . \mathrm { Cl } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { HCl } + \mathrm { HClO }$ is a reversible reaction, and $\mathrm { Cl } _ { 2 }$ does not react completely, then $\mathrm { Cl } _ { 2 }$ molecules are still present in the solution, and the sum of the number of $\mathrm { Cl } ^ { - } , \mathrm { HClO } , \mathrm { ClO } ^ { - }$'s in solution is less than the number of error; B. is a liquid in the standard condition and cannot be calculated in terms of the molar volume of the gas to calculate the number of nonpolar covalent bonds in it, B. Error ; C. The amount of substance of 56.0 g of Fe is 1 mol . The oxidizing property of $\mathrm { Br } _ { 2 }$ is so strong that it can oxidize Fe to $\mathrm { Fe } ^ { 3 + }$, and the equation of the reaction is $3 \mathrm { Br } _ { 2 } + 2 \mathrm { Fe } \xlongequal { \text { ignite } } 2 \mathrm { FeBr } _ { 3 }$, which shows that $\mathrm { Br } _ { 2 }$ is not enough when the amount of the two substances is equal. When $\mathrm { Br } _ { 2 }$ is insufficient, the electron transfer is calculated according to the amount of $\mathrm { Br } _ { 2 }$, and the number of electrons transferred is $2 N _ { A }$ for the reaction of $2 N _ { A }$, which is correct; D. HF is a liquid under standard conditions, and its amount of substance cannot be calculated from the molar volume of the gas, nor can the number of protons it contains be calculated, D is wrong; Therefore, the reasonable choice is C. $40 . \mathrm { C }$ [Knowledge Points] $22.4 \mathrm {~L} / \mathrm { mol }$ Conditions of Application, Calculation of Avogadro's Constant, Basic Calculation of Concentration of Substances [Detailed Explanation]A. $\mathrm { N } _ { \mathrm { A } }$ molecules of $\mathrm { O } _ { 2 }$ have an amount of matter of 1 mol, it is not clear whether it is the standard condition or not, and the volume is not necessarily 22.4 L, so A is wrong; B. Under standard conditions, ethanol is a liquid, so the quantity of 22.4 L of ethanol is not 1 mol, so B is wrong; C. At room temperature and pressure, the number of oxygen atoms in $16 \mathrm { gO } _ { 2 }$ is $\frac { 16 \mathrm {~g} } { 32 \mathrm {~g} / \mathrm { mol } } \times 2 \times \mathrm { N } _ { \mathrm { A } } = \mathrm { N } _ { \mathrm { A } }$, so C is correct; D. Without the volume of the solution, the amount of solute in the $0.5 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ solution of $\mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 }$ cannot be calculated, so D is wrong.

Question 38: 40. Using $\mathrm { N } _ { \mathrm { A } }$ to represent the value of Avogadro's constant, one of ...

40. Using $\mathrm { N } _ { \mathrm { A } }$ to represent the value of Avogadro's constant, one of the following statements is correct High School Chemistry Assignment, October 30, 2025

  • A. A. The $\mathrm { N } _ { \mathrm { A } }$ molecules of $\mathrm { O } _ { 2 }$ occupy a volume of about 22.4 L
  • B. B. The number of carbon atoms in 22.4 L of ethanol under standard conditions is $2 \mathrm {~N} _ { \mathrm { A } }$
  • C. C. The number of oxygen atoms in $16 \mathrm { gO } _ { 2 }$ is $\mathrm { N } _ { \mathrm { A } }$ at room temperature and pressure.
  • D. D. The $0.5 \mathrm {~mol} \cdot \mathrm {~L} ^ { - 1 }$ solution of $\mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 }$ contains $\mathrm { Na } ^ { + }$ in $\mathrm { N } _ { \mathrm { A } }$ number.

Answer: C

Solution:
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Mole Calculations

物质的量计算

38 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Calculating the amount of a substance is a central tool in quantitative chemical analysis, involving conversions between mass, volume, concentration and number of particles. In the CSCA exam, this knowledge is often tested in combination with chemical reactions, gas laws and solution preparation, and requires candidates to be able to skillfully apply the mole concept to solve practical problems. Questions usually require multi-step calculations and unit conversions.

Questions:38

Key Points

  • 1Conversion between mass, volume and quantity of matter
  • 2Quantitative relationship between reactants and products in chemical reactions
  • 3Molar volume of gases in standard conditions
  • 4Calculation of solution concentration (mol/L) and amount of solute substance

Study Tips

It is recommended that quantitative relationships in multi-step reactions be sorted out by drawing a flow chart of the calculations, always paying attention to the harmonization of units.

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