JEE Advance - Chemistry (2009 - Paper 1 Offline - No. 5)
The 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 0.8. The number of moles of N$$_2$$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
4.0 $$\times$$ 10$$^{-4}$$
4.0 $$\times$$ 10$$^{-5}$$
5.0 $$\times$$ 10$$^{-4}$$
4.0 $$\times$$ 10$$^{-6}$$
Explanation
According to Henry's law, we have
$${p_{{N_2}}} = {K_H}{x_{{N_2}}}$$
$$0.8 \times 5 = 1 \times {10^5} \times {x_{{N_2}}}$$
$${x_{{N_2}}} = 4 \times {10^{ - 5}}$$
Thus one mole of solution contains $$4 \times {10^{ - 5}}$$ moles of N$$_2$$ and $$1 - 4 \times {10^{ - 5}} \approx 1$$ mole of water. Therefore, N$$_2$$ from air dissolved in 10 moles of water is $$10 \times 4 \times {10^{ - 5}}$$ moles $$ = 4 \times {10^{ - 4}}$$.
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