Question

# The Henry's law constant for the solubility of $$N_2$$ gas in water at $$298\:K$$ is $$1\times 10^{\displaystyle5}\$$atm. The mole fraction of $$N_2$$ in air is $$0.8$$. Calculate the number of moles of $$N_2$$ dissolved in $$10\:moles$$ of water at $$298\:K$$ and $$5\:atm$$.

A
5×104
B
3×105
C
4×104
D
4×105

Solution

## The correct option is D $$4\times 10^{\displaystyle-4}$$Given,$$P_t$$ = $$5$$ atm. $$X_{N_{2}}$$ =  $$0.8$$$$K_H$$ = $$1\times 10^5$$ atmThe partial pressure of nitrogen is $$0.8 \times 5 = 4\ atm$$According to Henry's law, $$P=K_H\times X$$Substituting values in the above expression, we get - $$X = \dfrac {4 atm} {1 \times 10^{5}atm}=4 \times 10^{-5}\$$1 mole of water contains $$4\times 10^{-5}$$ moles of $$N_2$$.$$\therefore \text 10\ moles\ of\ water\ contains\ 4.0 \times 10^{-4}\ mol$$ Hence, option $$C$$ is correct.Chemistry

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