Question

The radius of an atom of He is 0.05 nm. Assuming that one mole of a gas occupies 22.4 litres at

STP, the fraction of the volume occupied by the atoms in a mole of He gas at STP is:

A. 1.4 × 10−4

. B. 1.4 × 10−5

. C. 7.1 × 10−4

. D. 7.1 × 10−5

.

Answer 1

Final. Answer :
$1.4 \times {10}^{ - 5}$
Steps and Understanding :
1) Radius of Helium Atom : r = 0.05 nm

$r = 5 \times {10}^{ - 11} m$

Volume of Helium atom, V
$= \frac{4}{3} \pi {r}^{3}$

2) Volume of Helium gas in any substance /container = 22.4 L

$= 22.4 \times {10}^{ - 3} {m}^{3}$

3) No. of Helium atoms in 1 mole He gas,
$n => 6.022 \times {10}^{23}$

4) Required Fraction,F =
$\frac{n \times \frac{4}{3} \pi {r}^{3} }{22.4 \times {10}^{ - 3} } \\ = \frac{6.022 \times {10}^{23} \times 4 \times \pi \times {(5 \times {10}^{ - 11} ) }^{3} }{22.4 \times {10}^{ - 3} \times 3} \\ \\ = > \frac{125 \times 6.022 \times 4 \times \pi \times {10}^{ - 7} }{22.4 \times 3} \\ \\ = > 140.76 \times {10}^{ - 7} \\ = > 1.4 \times {10}^{ - 5}$
Hence, Required Fraction is
$1.4 \times {10}^{ - 5}$