Solutions and Colligative Properties 1 Question 9
10. The Henry’s law constant for the solubility of $\mathrm{N}{2}$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^{5} \mathrm{~atm}$. The mole fraction of $\mathrm{N}{2}$ in air is 0.8 . The number of moles of $\mathrm{N}_{2}$ from air dissolved in 10 moles of water of $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is
(2009)
(a) $4.0 \times 10^{-4}$
(b) $4.0 \times 10^{-5}$
(c) $5.0 \times 10^{-4}$
(d) $4.0 \times 10^{-6}$
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Solution:
- Give, $K_{\mathrm{H}}=1 \times 10^{5} \mathrm{~atm}, \chi_{\mathrm{N}_{2}}=0.8$
$n_{\mathrm{H}{2} \mathrm{O}}=10$ moles, $p{\text {total }}=5 \mathrm{~atm}$
$p_{\mathrm{N}{2}}=p{\text {total }} \times \chi_{\mathrm{N}_{2}}=5 \times 0.8=4 \mathrm{~atm}$
According to Henry’s law,
$$ p_{\mathrm{N}{2}}=K{\mathrm{H}} \times \chi_{\mathrm{N}_{2}} $$
$$ \begin{aligned} 4 & =10^{5} \times \chi_{\mathrm{N}{2}} \ \chi{\mathrm{N}{2}} & =4 \times 10^{-5} \ \frac{n{\mathrm{N}{2}}}{n{\mathrm{N}{2}}+n{\mathrm{H}{2} \mathrm{O}}} & =4 \times 10^{-5} \ \frac{n{\mathrm{N}{2}}}{n{\mathrm{N}{2}}+10} & =4 \times 10^{-5} \ n{\mathrm{N}_{2}} & =4 \times 10^{-4} \end{aligned} $$
