Solutions and Colligative Properties - Result Question 9
10. 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 of $298 K$ and $5$ $atm$ 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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Answer:
Correct Answer: 10. (a)
Solution:
Give, $K _H=1 \times 10^{5} atm, \chi _{N _2}=0.8$
$n _{H _2 O}=10$ moles, $p _{\text {total }}=5 $ $atm$
$p _{N _2}=p _{\text {total }} \times \chi _{N _2}=5 \times 0.8=4 $ $atm$
According to Henry’s law,
$ p _{N _2}=K _H \times \chi _{N _2} $
$ \begin{aligned} 4 & =10^{5} \times \chi _{N _2} \\ \chi _{N _2} & =4 \times 10^{-5} \\ \frac{n _{N _2}}{n _{N _2}+n _{H _2 O}} & =4 \times 10^{-5} \\ \frac{n _{N _2}}{n _{N _2}+10} & =4 \times 10^{-5} \\ n _{N _2} & =4 \times 10^{-4} \end{aligned} $
BETA
