Question

The Henry's law constant for the solubility of $N_2$ gas in water at $298K$ is $1\times {10}^5$atm. The mole fraction of $N_2$ in air is $0.8$. Calculate the number of moles of $N_2$ dissolved in $10moles$ of water at $298K$ and $5atm$.

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

Answer 1

Answer: (C)

$4\times {10}^{-4}$

Given,

$P_t$ = $5$ atm.

$X_{N_2}$ = $0.8$

$K_H$ = $1\times {10}^5$ atm

The partial pressure of nitrogen is $0.8\times 5=4atm$

According to Henry's law, $P=K_H\times X$

Substituting values in the above expression, we get -

$X=\frac{4atm}{1\times {10}^{5}atm}=4\times {10}^{-5}$

1 mole of water contains $4\times {10}^{-5}$ moles of $N_2$.

$\therefore \text{1}0molesofwatercontains4.0\times {10}^{-4}mol$

Hence, option $C$ is correct.