Question

The vapour pressure of liquid water is 29.8 torr at ${25}^{\circ }\text{C}$. By what factor does the molar volume of water increases as it vaporizes to form an ideal gas under these conditions ? The density of liquid water is $1{\text{g cm}}^{-3}$ at $298\text{K}$.

• A. $3.46\times {10}^6$
• B. $2.31\times {10}^{-5}$
• C. $3.46\times {10}^4$
• D. $2.31\times {10}^{-7}$

Answer 1

The correct option is C $3.46\times {10}^4$

Molar volume of liquid water $\left(V_l\right)=\frac{\text{Mass}}{\text{Density}}$
$=\frac{18\text{g /mol}}{1\text{g}/c{m}^3}$
$=\frac{18}{1\times {10}^3}L/mol$ $(\because 1c{m}^3=\frac{1}{1000}L)$
According to the question, water vapours follow ideal gas equation. Hence molar volume of vapours is
$\left(V_g\right)=\frac{RT}{P}$
$=\frac{0.0821\text{L atm}{\text{K}}^{-1}\times 298K}{\left(\frac{29.8}{760}\text{atm}\right)}$
Ratio of molar volume of vapour and liquid solution =$\frac{V_g}{V_l}=\frac{0.0821\times 298\times 1\times 1000}{\left(\frac{29.8}{760}\right)\times 18}=3.46\times {10}^4$