Question

The Henry's law constant for the solubility of $N_2$ gas in water at 298 K is $1\times {10}^5$ atm. The mole fraction of $N_2$ in air is 0.8. The number of moles of $N_2$ from a dissolved in 100 moles of water at 298 K and 5 atm pressure is:

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

Answer 1

Answer: (C)

$4\times {10}^{-3}$

$P_{N_2}=K_H{X}_{N_2}$

$0.8\times 5=1\times {10}^5\times X_{N_2}$

$X_{N_2}=4\times {10}^{-5}$ (in 100 moles of water)

$=4\times {10}^{-5}=\frac{n_{N_2}}{n_{N_2}+100}$

$n_{N_2}\times 4\times {10}^{-5}+4\times {10}^{-3}=n_{N_2}$

$n_{N_2}\approx 4\times {10}^{-3}$