The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: $1\ \text{Å} = 10^{-10}\ \text{m}$. The size of a hydrogen atom is about $0.5\ \text{Å}$. What is the total atomic volume in $\text{m}^3$ of a mole of hydrogen atoms?

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1Å = 10⁻¹⁰m. Th

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1\ \text{Å} = 10^{-10}\ \text{m}

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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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Treating each hydrogen atom's given $0.5\ \text{Å}$ size as its radius, one atom occupies about $5.24\times10^{-31}\ \text{m}^3$ ; scaled up by Avogadro's number, one mole of hydrogen atoms occupies a total volume of about $\boxed{3.15\times10^{-7}\ \text{m}^3}$ .

Reading the given size correctly

The problem states the size of a hydrogen atom is $0.5\ \text{Å}$ . In this NCERT problem, this value is taken as the atom's radius (the standard atomic-radius quantity used in this kind of estimate), not its diameter:

r = 0.5\ \text{Å} = 0.5\times10^{-10}\ \text{m} = 5\times10^{-11}\ \text{m}

Volume of one hydrogen atom (as a sphere)

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

r^3 = (5\times10^{-11})^3 = 125\times10^{-33} = 1.25\times10^{-31}\ \text{m}^3

V_{\text{atom}} = \frac{4}{3}\pi \times 1.25\times10^{-31} \approx 5.24\times10^{-31}\ \text{m}^3

Scaling up to one mole

One mole contains $N_A = 6.022\times10^{23}$ atoms: …

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Common Mistakes: Atomic Volume of a Mole of Hydrogen Atoms

This question gives the hydrogen atom's size as 0.5 Å and, following NCERT's own treatment, this is taken directly as the atom's radius (not its diameter) -- so $r = 0.5\ \text{Å} = 5\times10^{-11}\ \text{m}$ . Getting this one interpretation right or wrong changes the final answer by a factor of 8 (since volume scales as $r^3$ ), so it's the single most important thing to get right on this problem.

Mistake 1: Treating 0.5 Å as a diameter and halving it again

The error: Reading "size... is about 0.5 Å" as a diameter, then computing radius $= 0.25\ \text{Å} = 2.5\times10^{-11}\ \text{m}$ .

Why it's wrong: For this specific NCERT problem, the given 0.5 Å is the radius -- that is the standard, textbook-prescribed reading. Using it as a diameter shrinks the radius by half and the volume by a factor of $2^3 = 8$ , giving a final answer of roughly $3.9\times10^{-8}\ \text{m}^3$ instead of the correct $\approx 3.15\times10^{-7}\ \text{m}^3$ .

How to avoid: Take $r = 0.5\ \text{Å} = 5\times10^{-11}\ \text{m}$ directly, with no extra halving step.

Mistake 2: Forgetting to convert Å to metres before cubing

The error: Plugging $r = 0.5$ (or $5$ ) straight into $V = \tfrac{4}{3}\pi r^3$ without first converting to SI units, since the answer is required in $\text{m}^3$ .

How to avoid: Always write the conversion explicitly first: $1\ \text{Å} = 10^{-10}\ \text{m}$ , so $r = 0.5\times10^{-10}\ \text{m} = 5\times10^{-11}\ \text{m}$ , then cube it.

Mistake 3: Forgetting to scale up to one mole

The error: Computing the volume of a single hydrogen atom ( $\approx 5.24\times10^{-31}\ \text{m}^3$ ) and stopping there -- the question asks for the volume of a mole of atoms. …

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←Previous question1.13 — A famous relation in physics relates 'moving mass' m to the 'rest mass…→Next question1.15 — One mole of an ideal gas at standard temperature and pressure occupies…

Q.The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: $1\ \text{Å} = 10^{-10}\ \text{m}$ . The size of a hydrogen atom is about $0.5\ \text{Å}$ . What is the total atomic volume in $\text{m}^3$ of a mole of hydrogen atoms?

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

Reading the given size correctly

Volume of one hydrogen atom (as a sphere)

Scaling up to one mole

Working exponent and mantissa separately

Common Mistakes: Atomic Volume of a Mole of Hydrogen Atoms

Mistake 1: Treating 0.5 Å as a diameter and halving it again

Mistake 2: Forgetting to convert Å to metres before cubing

Mistake 3: Forgetting to scale up to one mole

Reading the given size correctly

Volume of one hydrogen atom (as a sphere)

Scaling up to one mole

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