Question

One kg of water under a nitrogen pressure of 1 atm dissolves $0.02of{N}_{2}at293K$. Calculate Henry's law constant.

• A. $7.75\times {10}^2{atm}^{-1}$
• B. $7.75\times {10}^{-4}{atm}^{-1}$
• C. $1.27\times {10}^4{atm}^{-1}$
• D. $7.75\times {10}^5{atm}^{-1}$

Answer 1

The correct option is B $7.75\times {10}^{-4}{atm}^{-1}$

$\begin{array}{c}\text{For water}1Kg=1L\\ 0.02g\text{nitrogen corresponds}t0=\frac{0.02}{28}=7.2\times {10}^{-4}moles\end{array}$

Hence, the solubility $4s=7.2\times {10}^{-4}M$

Accrding to Henry. Law $s=kp$

$\begin{array}{c}7\cdot 2\times {10}^{-4}=k\times 1atm\\ K=7.2\times {10}^{-4}{atm}^{-1}\end{array}$

Correct Answer=B