Ncert Exercises & Solutions

One mole of an ideal gas at standard temperature and pressure occupies 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 the hydrogen molecule to be about 1 Å). Why is this ratio so large?

Radius of hydrogen atom, r = 0.5 Å = 0.5 × $10^{- 10}$ m

Volume of hydrogen atom = $\frac{4}{3} {πr}^{3}$

$= \frac{4}{3} \times \frac{22}{7} \times {(0.5 \times 10^{- 10})}^{3}$
$= 0.524 \times 10^{- 30} m^{3}$

Now, 1 mole of hydrogen contains 6.023 × $10^{23}$ hydrogen atoms.

$\Rightarrow$ The volume of 1 mole of hydrogen atoms,

V_{a} = 6.023 \times 10^{23} \times 0.524 \times 10^{- 30} = 3.16 \times 10^{- 7} m^{3}

Molar volume of 1 mole of hydrogen atoms at STP,

$V_{m} = 22.4 L = 22.4 \times 10^{- 3} m^{3}$
$\Rightarrow \frac{V_{m}}{V_{a}} = \frac{22.4 \times 10^{- 3}}{3.16 \times 10^{- 7}} = 7.08 \times 10^{4}$

Hence, the molar volume is $7.08 \times 10^{4}$ times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

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