Question

In a mixture of carbon dioxide, nitrogen and water
vapour, the mole fraction of nitrogen is 0.8. The
number of moles of nitrogen from mixture dissolved
in 10 mol of water at 298 K and 8 atm is
(Henry's constant of N, gas in water at 298 K is
10^5atm)
(1) 4.8 * 104
(2) 6.4 x 104
(3) 5.0 * 104
(4) 4.8 * 10-​

Answer 1

answer : option (2) 6.4 × 10^-4 mol

explanation : mole fraction of nitrogen = 0.8

pressure of mixture, P = 8atm.

so, pressure of nitrogen in mixture = 0.8 × 8 atm = 6.4 atm.

from Henry's law

$P_{N_2}=K_Hx_{N_2}$

where $P_{N_2}$ is pressure of nitrogen in mixture, $K_H$ is Henry's constant and $x_{N_2}$ is mole fraction of nitrogen in 10 mol of water at 298K.

or, 6.4 = 10^5 × $x_{N_2}$

or, $x_{N_2}$ = 6.4 × 10^-5

so, mole fraction of nitrogen in 10mol of water , $x_{N_2}=\frac{n_{N_2}}{n_{N_2}+10}$

or, 6.4 × 10^-5 = $\frac{n_{N_2}}{n_{N_2}+10}$

or, $n_{N_2}$ ≈ 6.4 × 10^-4 mol

hence, option (2) is correct choice.