Henry's law constants for O_2 and N_2 are K_{ | vedclass

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(A) According to Henry's law,the partial pressure of a gas is given by $p = K_H \cdot X$,where $K_H$ is the Henry's law constant and $X$ is the mole f

Vedclass · Accepted Answer

$0.49$

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(A) According to Henry's law,the partial pressure of a gas is given by $p = K_H \cdot X$,where $K_H$ is the Henry's law constant and $X$ is the mole fraction.For $O_2$: $p_{O_2} = K_{O_2} \cdot X_{O_2} \implies X_{O_2} = \frac{p_{O_2}}{K_{O_2}}$For $N_2$: $p_{N_2} = K_{N_2} \cdot X_{N_2} \implies X_{N_2} = \frac{p_{N_2}}{K_{N_2}}$The ratio of mole fractions is $\frac{X_{O_2}}{X_{N_2}} = \frac{p_{O_2}}{K_{O_2}} 	imes \frac{K_{N_2}}{p_{N_2}} = \left(\frac{p_{O_2}}{p_{N_2}}ight) 	imes \left(\frac{K_{N_2}}{K_{O_2}}ight)$.Air contains $20 \%$ $O_2$ and $80 \%$ $N_2$ by volume,so the ratio of partial pressures is $\frac{p_{O_2}}{p_{N_2}} = \frac{20}{80} = \frac{1}{4} = 0.25$.Substituting the values: $\frac{X_{O_2}}{X_{N_2}} = 0.25 	imes \frac{6.51 	imes 10^7}{3.3 	imes 10^7} = 0.25 	imes 1.9727 \approx 0.493$.

Similar Questions

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State Henry's law and explain it. Also,mention its applications.

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As temperature increases,the value of $K_H$ increases,thus the value of solubility of a gaseous solute will $ . . . . . . $.

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