Question

Air contains O2 nd N2 in the ratio 0.2 :0.8 if Henry's law constant for O2 and N2 are 3.3 ×10^7 torr and 6.6×10^7 torr respectively, then the ratio of mole fractions of O2 and N2 dissolved in water at 1 bar pressure is ??​

Answer 1

Answer:

1:2

Explanation:

given

3.3×(10)^7:6.6×10^7

1:2

Answer 2

The ratio of mole fractions of oxygen gas and nitrogen gas is 1 : 8

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

where,

$K_H$ = Henry's constant

$C_{A}$ = molar solubility of gas A

$p_{A}$ = partial pressure of gas A

The molar solubilities of the gases are in terms of mole fraction in the solution

Taking the ratios of mole fraction of the gases:

$\frac{\chi_{O_2}}{\chi_{N_2}}=\frac{K_H(O_2)\times p_{O_2}}{K_H(N_2)\times p_{N_2}}$

We are given:

$\frac{p_{O_2}}{p_{N_2}}=\frac{0.2}{0.8}=\frac{1}{4}$

$K_H(O_2)=3.3\times 10^7torr$

$K_H(N_2)=6.6\times 10^7torr$

Putting values in above expression, we get:

$\frac{\chi_{O_2}}{\chi_{N_2}}=\frac{3.3\times 10^7}{6.6\times 10^7}\times \frac{1}{4}\\\\\frac{\chi_{O_2}}{\chi_{N_2}}=\frac{1}{8}$

Learn more about Henry's law:

https://brainly.com/question/14735276

https://brainly.com/question/14846824

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