Question

Henry's law constant for the solubility of $N_2$ gas in water at $298$ K is $1.0\times {10}^5$ atm. The mole fraction of $N_2$ in air is $0.6$. The number of moles of $N_2$ from air dissolved in $10$ moles of water at $298$ K and $5$ atm pressure is:

• A. $3.0\times {10}^{-4}$
• B. $4.0\times {10}^{-4}$
• C. $5.0\times {10}^{-4}$
• D. $4.0\times {10}^{-5}$

Answer 1

Answer: (A)

$3.0\times {10}^{-4}$

Acc. to Henry's law

$X_a\left(g\right)\propto P_{(}g)$ and $X_a\times K_h=P_g$

So the partial pressure of gas $=P_{a}tm\times X_{N_2}$

$=5\times .6=3atm$

Now $P_g=K_h\times X_g$

$3={10}^5\times \frac{n}{10}$

$n=3\times {10}^{-4}mol$