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Standard XII

Chemistry

Solids in Liquids

The Henry's l...

Question

The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 10^{5} a t m$ . 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 $5$ atmospheric pressure is $x \times 10^{- y}$ . Find the value of $\frac{x}{y}$ .

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Solution

According to Henry's law,
$x_{N_{2}} \times K_{H} = p_{N_{2}}$
$(p_{N_{2}} = Partial pressure of N_{2})$
Given,
Total pressure $= 5 a t m$
Mole fraction of $N_{2} = 0.8$
$∴ Partial pressure of N_{2} = 0.8 \times 5 = 4 \Rightarrow x_{N_{2}} \times 1 \times 10^{5} = 4 \Rightarrow x_{N_{2}} = 4 \times 10^{- 5} Number of moles of H_{2} O, n_{H_{2} O} = 10 Number of moles of N_{2}, n_{N_{2}} = ? \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2} O}} = x_{N_{2}} = 4 \times 10^{- 5} \frac{n_{N_{2}}}{10 + n_{H_{2} O}} = x_{N_{2}} = 4 \times 10^{- 5} \Rightarrow n_{N_{2}} = 4 \times 10^{- 4} H e n c e, \frac{x}{y} = \frac{4}{4} = 1$

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The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0×105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atmospheric pressure is x×10−y. Find the value of xy.