物质的量wùzhì de liàng

Core Concept

The mole (物质的量) is the SI unit for amount of substance. One mole contains Avogadro's number of particles:

$N_A = 6.02 \times 10^{23}\ \mathrm{mol^{-1}}$

A point of precision the exam checks: 物质的量 is the quantity, and mol is its unit — just as length is a quantity and metre is its unit. Saying "the 物质的量 is 2 mol" is correct; saying "the mole is 2" is not.

Always state the particle. "1 mol of hydrogen" is ambiguous — $\mathrm{H}$ atoms or $\mathrm{H_2}$ molecules? Write $1\ \mathrm{mol\ H_2}$.

The Conversion Hub

The mole is the junction every stoichiometry calculation passes through:

$n = \frac{m}{M} = \frac{N}{N_A} = \frac{V}{V_m} = c \cdot V_{\text{solution}}$

Symbol	Quantity	Unit
$n$	amount of substance	$\mathrm{mol}$
$m$	mass	$\mathrm{g}$
$M$	molar mass	$\mathrm{g/mol}$
$N$	number of particles	—
$V$	gas volume	$\mathrm{L}$
$V_m$	molar volume of gas	$\mathrm{L/mol}$
$c$	concentration	$\mathrm{mol/L}$

Mass never converts directly to particle count — it goes through $n$. Drawing that hub once fixes most calculation errors.

Molar Volume

At STP (标准状况: $0\ \mathrm{^\circ C}$, $101\ \mathrm{kPa}$):

$V_m = 22.4\ \mathrm{L/mol}$

Three conditions on this number, all examined:

1. Gases only — never for liquids or solids
2. STP only — not room temperature
3. Applies to any gas or gas mixture, since it depends on particle count, not identity

Concentration

$c = \frac{n}{V_{\text{solution}}}$

$V$ is the volume of the solution, not of the solvent added. Dissolving something increases the volume, so the two differ.

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★☆☆☆)

How many moles are in $9.8\ \mathrm{g}$ of $\mathrm{H_2SO_4}$? ($M = 98\ \mathrm{g/mol}$)

Options:

• A. $0.05\ \mathrm{mol}$
• B. $0.1\ \mathrm{mol}$
• C. $1\ \mathrm{mol}$
• D. $10\ \mathrm{mol}$

Detailed Solution:

$n = \frac{m}{M} = \frac{9.8}{98} = 0.1\ \mathrm{mol}$

Option D inverts the fraction. A quick sanity check: $9.8\ \mathrm{g}$ is about a tenth of $98\ \mathrm{g}$, so the answer must be about a tenth of a mole.

Answer: B

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Example 2: Intermediate (Difficulty ★★★☆☆)

What volume of $\mathrm{H_2}$ at STP is produced when $6.5\ \mathrm{g}$ of $\mathrm{Zn}$ reacts completely with excess $\mathrm{HCl}$? ($M_{\mathrm{Zn}} = 65\ \mathrm{g/mol}$)

Options:

• A. $1.12\ \mathrm{L}$
• B. $2.24\ \mathrm{L}$
• C. $4.48\ \mathrm{L}$
• D. $22.4\ \mathrm{L}$

Detailed Solution:

1. Write the equation: $\mathrm{Zn + 2HCl \rightarrow ZnCl_2 + H_2\uparrow}$

2. Convert mass to moles: $n(\mathrm{Zn}) = \frac{6.5}{65} = 0.1\ \mathrm{mol}$

3. Use the mole ratio from the equation — $\mathrm{Zn : H_2} = 1 : 1$: $n(\mathrm{H_2}) = 0.1\ \mathrm{mol}$

4. Convert moles to gas volume at STP: $V = n \cdot V_m = 0.1 \times 22.4 = 2.24\ \mathrm{L}$

Answer: B

The pattern to internalise: mass → mol → mole ratio → mol → target quantity. Every stoichiometry question is that chain; only the endpoints change.

Common Mistakes

❌ Mistake 1: Applying $22.4\ \mathrm{L/mol}$ to liquids or solids

$1\ \mathrm{mol}$ of water is $18\ \mathrm{mL}$, not $22.4\ \mathrm{L}$.

❌ Mistake 2: Using $22.4$ away from STP

At room temperature the molar volume is about $24\ \mathrm{L/mol}$. If the question does not say 标准状况, be careful.

❌ Mistake 3: Skipping the mole ratio

Converting reactant mass to moles and then straight to product volume ignores the equation's coefficients.

❌ Mistake 4: Using solvent volume as solution volume

$c = n/V_{\text{solution}}$. Adding $1\ \mathrm{L}$ of water does not give a $1\ \mathrm{L}$ solution.

Study Tips

1. ✅ Draw the conversion hub with $n$ at the centre before every calculation
2. ✅ Always route through moles — never mass directly to volume
3. ✅ Check the three conditions before writing $22.4$
4. ✅ Name the particle whenever you write a mole quantity
5. ✅ Estimate first — is the answer bigger or smaller than one mole?

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💡 Exam Tip: 化学方程式中物质的量的运用 and 物质的量浓度的基础计算 are both frequent CSCA knowledge points. The mole is the hub every quantitative chemistry question routes through, so fluency here pays across the whole paper.