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
B
4.0×104
C
5.0×104
D
4.0×105

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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