At 300 K, a 0.5 mole sample of air (80% by mol of N₂ & 20% by mol of O₂) is put in 10 litre cylinder piston system in which 5 litre water is already present, then which of the following is/are correct: (R = 0.0821 lt atm / (K-mol)
k_H for $N_{2} = 10^{5} atm$
k_H for $O_{2} = 4 \times 10^{4} atm$

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$X_{O_{2}}$ in $H_{2} O$ will be $1.2315 \times 10^{- 5}$

On increasing temperature value of Henry’s constant increases

$X_{N_{2}}$ in $H_{2} O$ will be $1.9704 \times 10^{- 5}$

In $H_{2} O$ , $N_{2} (g)$ is more soluble as compare to $O_{2} (g)$ at same pressure

answer is A, B, C.

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Detailed Solution

$P_{N_{2}} = \frac{n_{H_{2}} RT}{V} = \frac{0 .5 \times 0 .8 \times 0 .0821 \times 300}{5} = 0 .8 \times 0 .0821 \times 30 P_{N_{2}} = {(K_{H})}_{N_{2}} x_{N_{2}} x_{N_{2}} = 0.8 \times 0.082 \times 30 \times 10^{- 5} x_{N_{2}} = 1.9704 \times 10^{- 5} P_{O_{2}} = \frac{0.5 \times 0.2 \times 0.0821 \times 300}{5} = 0.2 \times 0.821 \times 3 x_{O_{2}} = \frac{P_{O_{2}}}{{(k_{H})}_{O_{2}}} = \frac{0.2 \times 0.821 \times 3}{4 \times 10^{+ 4}} = 1.2315 \times 10^{- 5}$