CO_2 gas is bubbled through water during a so | vedclass

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(B) According to Henry's Law,$P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas.Given: $P = 0.835 \ bar$,$K_H = 1.67 	imes 10^3 \ bar

Vedclass · Accepted Answer

$25$

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(B) According to Henry's Law,$P = K_H 	imes \chi$,where $\chi$ is the mole fraction of the gas.Given: $P = 0.835 \ bar$,$K_H = 1.67 	imes 10^3 \ bar$,Volume of water = $0.9 \ L = 900 \ g$.Moles of water $(n_{H_2O})$ = $\frac{900 \ g}{18 \ g/mol} = 50 \ mol$.Since the amount of dissolved gas is very small,$\chi_{CO_2} \approx \frac{n_{CO_2}}{n_{H_2O}}$.$0.835 = 1.67 	imes 10^3 	imes \frac{n_{CO_2}}{50}$.$n_{CO_2} = \frac{0.835 	imes 50}{1.67 	imes 10^3} = \frac{41.75}{1670} = 0.025 \ mol$.$x \ mmol = 0.025 	imes 1000 = 25 \ mmol$.Therefore,the value of $x$ is $25$.

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