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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
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B
3×105
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C
4×104
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D
4×105
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Solution

The correct option is D $$4\times 10^{\displaystyle-4}$$
Given,

$$P_t$$ = $$5$$ atm. 
$$X_{N_{2}}$$ =  $$0.8$$
$$K_H$$ = $$1\times 10^5$$ atm

The 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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