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Air contains O₂ and N₂ in the ratio 0.2 : 0.8. If Henry law constant for O₂ and N₂ are 3.3 × 10⁷torr and 6.6 × 10⁷ torr respectively, then the ratio of mole fractions of O₂ to N₂ dissolved in water at 1 bar pressure is

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1 : 1

2 :1

1 : 2

1 : 3

answer is C.

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Detailed Solution

mole ratio of O₂ : N₂ in air =0.2 : 0.8 = 1:4

We know

$large begin{array}{l} {P_A} = {P_{total}}{X_A}\ {P_A} propto {X_A}\ Thus,,,{P_{{O_2}}}:{P_{{N_2}}} = 1:4 end{array}$

Given

$large K_{O_2}:K_{N_2}=(3.3times 10^7):(6.6times 10^7)=1:2$

According to Henry's law

$large P_A=K_HX_A$

where X_A = molefraction of gas dissolved in liquid

$large frac{P_{O_2}}{P_{N_2}}=frac{K_{O_2}times X_{O_2}}{K_{N_2}times X_{N_2}}$

$large frac{1}{4}=frac{1}{2}times frac{X_{O_2}}{X_{N_2}}$

$large frac{X_{O_2}}{X_{N_2}}=frac{1}{2}$

$large {X_{O_2}}:{X_{N_2}}={1}:{2}$

Thus ratio of molefraction of O₂ to N₂ dissolved in water is 1:2

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