Question

Given $H_{2}O(l)\rightleftharpoons H_{2}O(g),23.8$ torr at $25^{\circ }C$
Density of $H_{2}O(l)=1.0g/c{m}^3$ at $25^{\circ }C$
Thus, ratio of molar volume of water vapour to that of liquid water at $25^{\circ }C$ is

• A. 43.4
• B. $2.30\times 10^{-5}$
• C. $4.34\times 10^4$
• D. $2.41\times 10^3$

Answer 1

Answer: (C)

Density of water at $25^{\circ }C=1g/c{m}^3$

Thus, volume of 1 mole $H_{2}O$ (l) at $25^{\circ }C$

$=\frac{\text{ mass }}{\text{ density }}=\frac{18g}{1g/c{m}^3}=18c{m}^3$

Volume of 1 mole of water vapour at $25^{\circ }C=\frac{nRT}{p}$

$=\frac{1mol\times 0.0821Latmmo{l}^{-1}{K}^{-1}\times 298K}{\left(\frac{23.8}{760}\right)atm}$

$=781.26\times 10^{3}c{m}^3$

Thus, $\frac{V(\text{ vapour })}{V(\text{ liquid })}$

$=\frac{781.26\times 10^3}{18}$

$=4.34\times 10^4$