Question

Given $H _{2} O (l) \rightleftharpoons H _{2} O ( g ), 23.8$ torr at $25^{\circ} C$
Density of $H _{2} O (l)=1.0 g / cm ^{3}$ 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 $25^{\circ} C =1 \,g / cm ^{3}$

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

$=\frac{\text { mass }}{\text { density }}=\frac{18 \,g }{1\, g / cm ^{3}}=18\, cm ^{3}$

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

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

$=781.26 \times 10^{3} cm ^{3}$

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

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

$=4.34 \times 10^{4}$