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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
3.0×104
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B
4.0×104
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C
5.0×104
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D
4.0×105
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Solution

The correct option is C $$3.0\times 10^{-4}$$
Acc. to Henry's law
$$X_a(g) \propto P_(g)$$ and $$X_a \times K_h = P_g$$ 
So the partial pressure of gas $$= P_atm \times  X _{N_2}$$
                                    $$ = 5\times .6 = 3atm$$
Now $$P_g = K_h\times X_g$$
       $$ 3  = 10^5 \times \dfrac{n}{10}$$
       $$ n = 3\times 10^{-4} mol$$

Chemistry

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