One mole of an ideal gas at standard temperature and pressure occupies $22.4\ \text{L}$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about $1\ \text{Å}$). Why is this ratio so large?

One mole of an ideal gas at standard temperature and pressure occupies 22.4L (molar volume). What is the ratio

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22.4\ \text{L}

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Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Atoms are mostly empty space — a tiny dense nucleus wrapped in a fuzzy electron cloud. So "atomic volume" does not mean the volume of a solid ball; it means the average space one atom occupies when many atoms are packed together in a solid or liquid.

Think of a crowded hall: to find the space per person you divide the hall's volume by the number of people. Atomic volume does exactly that for atoms.

Definition

Atomic volume is the volume occupied by one mole of atoms of an element in its solid or liquid state. It is found from the element's molar mass and density:

V_{\text{atomic}} = \frac{M}{\rho}

$V_{\text{atomic}}$ = atomic volume (cm³/mol)
$M$ = molar mass (g/mol)
$\rho$ = density (g/cm³)

This gives the volume per mole. Dividing by Avogadro's number gives the space per single atom:

V_{\text{one atom}} = \frac{V_{\text{atomic}}}{N_A}, \qquad N_A = 6.022\times10^{23}\ \text{mol}^{-1}

Worked example — aluminium

Molar mass $M = 26.98$ g/mol, density $\rho = 2.70$ g/cm³:

V_{\text{atomic}} = \frac{26.98}{2.70} = 9.99\ \text{cm}^3/\text{mol}

So one mole of Al atoms occupies about 10 cm³. Per atom:

V_{\text{one Al atom}} = \frac{9.99}{6.022\times10^{23}} = 1.66\times10^{-23}\ \text{cm}^3

Periodic trends

Down a group: atomic volume increases — more electron shells make atoms larger.
Across a period: it generally decreases — rising nuclear charge pulls the electrons in tighter.
Allotropes differ: diamond is denser than graphite, so diamond's atomic volume is smaller, though both are carbon.

These trends help explain why alkali metals (large atomic volume) are soft and reactive, while transition metals (smaller atomic volume) are hard and dense. …

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

Atomic volume is not the volume of a single atom — it's the volume occupied by one mole of atoms of an element in its solid state. This is a macroscopic quantity that helps us understand how tightly atoms pack together.

The key formula is:

\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}

Let's break down why this works.

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Density ( $\rho$ ) is defined as:

\rho = \frac{\text{Mass}}{\text{Volume}}

For a pure solid element, if we take one mole of atoms:

Mass of one mole = Atomic mass (in g/mol)
Volume of one mole = Atomic volume (in cm³/mol)

So:

\rho = \frac{\text{Atomic mass}}{\text{Atomic volume}}

Step 2: Rearranging to find atomic volume

\text{Volume} = \frac{\text{Mass}}{\text{Density}}

Therefore:

\boxed{\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}}

Why This Makes Physical Sense

Atomic mass tells you how heavy one mole of atoms is; density tells you how much mass fits in a given space. Dividing mass by density gives the space that mass occupies.

Example intuition: If iron has atomic mass ≈ 56 g/mol and density 7.87 g/cm³, then:

\text{Atomic volume} = \frac{56}{7.87} \approx 7.1 \text{ cm}^3/\text{mol}

This means one mole of iron atoms (about $6.022 \times 10^{23}$ atoms) occupies roughly 7.1 cm³ of space.

Important Exam Points

Concept	Why It Matters
Units	Atomic mass in g/mol, density in g/cm³ → atomic volume in cm³/mol
Solid state only	The formula assumes atoms are closely packed; gases/liquids have different packing

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The ratio of molar volume to atomic volume is about $10^4$ , showing that gas molecules are mostly empty space — the molar volume is dominated by the vast gaps between molecules, not by the molecules themselves.

Why this approach works

The question asks for the ratio of two volumes: the molar volume (the space one mole of gas occupies) and the atomic volume (the actual space taken up by the molecules themselves). The huge ratio tells us something fundamental about gases: they are mostly empty space. At STP, molecules are far apart — the volume of the container is almost entirely empty, with the molecules themselves occupying only a tiny fraction.

We need to compare:

Molar volume: $22.4\ \text{L}$ (given)
Atomic volume: the total volume of all molecules in one mole

The "size of hydrogen molecule" is given as $1\ \text{Å}$ — this is the diameter. We treat each molecule as a sphere of that diameter, compute its volume, then multiply by Avogadro's number to get the total atomic volume for one mole.

Step-by-step solution

1. Understand the given data

Molar volume at STP: $V_m = 22.4\ \text{L} = 22.4 \times 10^{-3}\ \text{m}^3$
Diameter of hydrogen molecule: $d = 1\ \text{Å} = 1 \times 10^{-10}\ \text{m}$
Avogadro's number: $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

Watch out

A common mistake is to treat $1\ \text{Å}$ as the radius. The problem says "size" — in context, this means the diameter of the molecule. Always check: "size" of a molecule typically refers to its diameter unless stated otherwise.

2. Compute the volume of a single hydrogen molecule

Treat the molecule as a sphere. Volume of a sphere of radius $r$ is:

V_{\text{molecule}} = \frac{4}{3} \pi r^3

Radius $r = d/2 = 0.5 \times 10^{-10}\ \text{m} = 5 \times 10^{-11}\ \text{m}$ .

V_{\text{molecule}} = \frac{4}{3} \pi (5 \times 10^{-11})^3

First compute $(5 \times 10^{-11})^3 = 125 \times 10^{-33} = 1.25 \times 10^{-31}\ \text{m}^3$ .

Then:

V_{\text{molecule}} = \frac{4}{3} \pi \times 1.25 \times 10^{-31}

Using $\pi \approx 3.14$ :

\frac{4}{3} \times 3.14 \approx 4.19

So:

V_{\text{molecule}} \approx 4.19 \times 1.25 \times 10^{-31} = 5.24 \times 10^{-31}\ \text{m}^3

Tip

You can keep the expression symbolic: $V_{\text{molecule}} = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{\pi d^3}{6}$ . This is often faster. Here $d = 10^{-10}\ \text{m}$ , so $V_{\text{molecule}} = \frac{\pi}{6} \times 10^{-30}\ \text{m}^3 \approx 0.524 \times 10^{-30}\ \text{m}^3 = 5.24 \times 10^{-31}\ \text{m}^3$ . Same result.

3. Compute the atomic volume of one mole of hydrogen

Atomic volume = volume of one molecule × Avogadro's number:

V_{\text{atomic}} = V_{\text{molecule}} \times N_A

V_{\text{atomic}} \approx (5.24 \times 10^{-31}) \times (6.022 \times 10^{23})

Multiply:

5.24 \times 6.022 \approx 31.55

So:

V_{\text{atomic}} \approx 31.55 \times 10^{-8} = 3.155 \times 10^{-7}\ \text{m}^3

4. Compute the ratio

\text{Ratio} = \frac{V_m}{V_{\text{atomic}}} = \frac{22.4 \times 10^{-3}}{3.155 \times 10^{-7}}

Divide:

\frac{22.4}{3.155} \approx 7.10

And $10^{-3} / 10^{-7} = 10^4$ , so:

$\text{Ratio} \approx 7.10 \times 10^4$ …

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Common Mistakes: Ratio of Molar Volume to Atomic Volume

This question gives the hydrogen molecule's size as about 1 Å, and (following NCERT's own treatment) this is taken as the molecule's diameter, so the radius used in the volume formula is $r = 0.5\ \text{Å} = 5\times10^{-11}\ \text{m}$ -- the opposite reading convention from the neighbouring atomic-volume question, where the given size was taken directly as the radius. Mixing the two conventions up is the single most common error on this pair of questions.

Mistake 1: Treating 1 Å as the radius instead of the diameter

The error: Using $r = 1\ \text{Å} = 10^{-10}\ \text{m}$ directly, instead of first halving it.

Why it's wrong: Here the size given is the molecule's diameter, so the radius is half that: $r = 0.5\ \text{Å} = 5\times10^{-11}\ \text{m}$ . Skipping the halving step overstates the radius by 2x and the volume (and hence the atomic volume, and the final ratio) by a factor of 8.

How to avoid: Always check whether "size" in a given problem means radius or diameter -- here it's a diameter, so halve it before cubing.

Mistake 2: Mixing units between the two volumes being compared

The error: Comparing the molar volume in litres directly against the atomic volume in $\text{m}^3$ without converting both to the same unit first.

How to avoid: Convert everything to $\text{m}^3$ first: $22.4\ \text{L} = 22.4\times10^{-3}\ \text{m}^3$ , then take the ratio of the two volumes in that common unit.

Mistake 3: Forgetting to scale the molecule's volume up to one mole

The error: Computing only the volume of a single molecule and dividing the 22.4 L molar volume by that, instead of first multiplying by Avogadro's number to get the volume of a whole mole of molecules.

How to avoid: $V_{\text{atomic, mole}} = N_A \times V_{\text{molecule}}$ before taking the ratio with the molar (gas) volume.

Mistake 4: Leaving out the "why is it so large" explanation …

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Q.One mole of an ideal gas at standard temperature and pressure occupies $22.4\ \text{L}$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about $1\ \text{Å}$ ). Why is this ratio so large?

Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Definition

Worked example — aluminium

Periodic trends

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Step 2: Rearranging to find atomic volume

Why This Makes Physical Sense

Important Exam Points

Why this approach works

Step-by-step solution

Concept: Atomic Volume Calculation

Common Mistakes: Ratio of Molar Volume to Atomic Volume

Mistake 1: Treating 1 Å as the radius instead of the diameter

Mistake 2: Mixing units between the two volumes being compared

Mistake 3: Forgetting to scale the molecule's volume up to one mole

Mistake 4: Leaving out the "why is it so large" explanation …

Q.One mole of an ideal gas at standard temperature and pressure occupies 22.4 L22.4\ \text{L}22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 A˚1\ \text{Å}1 A˚). Why is this ratio so large?

Why this approach works

Step-by-step solution

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Q.One mole of an ideal gas at standard temperature and pressure occupies $22.4\ \text{L}$ (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about $1\ \text{Å}$ ). Why is this ratio so large?

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