Question

A gas 'X' is present with saturated water vapour over water liquid at total pressure of 1.5 atm. Vapour pressure of $H_{2}O$ of same temperature is 0.5 atm. what is the solubility of gas 'X' in term of moels in 10 mole $H_{2}O$ (l).

• A. 1 x 10${}^{-3}$ mole
• B. 5 x 10${}^{-2}$ mole
• C. 2 x 10${}^{-3}$ mole
• D. 1 x 10${}^{-2}$ mole

Answer 1

The correct option is C 2 x 10${}^{-3}$ mole

$\begin{array}{c}\text{Given slope}=38\times {10}^5\text{torr.}\\ \text{By Henry}law\\ \text{Amount of gas}\propto \text{partial pressure.}\end{array}$

$\begin{array}{c}{P}_T=1.5atm\\ P_{H_{2}O}=0.5atm\end{array}$

$\begin{array}{cc}{P}_T=& P_x+P_{H_{2}O}\\ P_{H_{2}O}=& 1\cdot 5-0.5\\ & =1atm\end{array}$

$P_x=k_H\cdot \frac{n{x}^{}}{n_{H_{2}O}}$

$\begin{array}{cc}1=& \frac{38\times {10}^5}{760}\times \frac{nx}{10}\\ nx& =2\times {10}^{-3}mole\end{array}$

$\text{so option}c\text{is correct}$