A company dissolves 'X' amount of CO_{2} at 2 | vedclass

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(D) According to Henry's Law: $P = K_{H} 	imes \chi_{CO_{2}}$Given:$P = 0.835 \ bar$$K_{H} = 1.67 \ kbar = 1670 \ bar$Since the amount of $CO_{2}$ is

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$1222$

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(D) According to Henry's Law: $P = K_{H} 	imes \chi_{CO_{2}}$Given:$P = 0.835 \ bar$$K_{H} = 1.67 \ kbar = 1670 \ bar$Since the amount of $CO_{2}$ is very small,we can approximate the mole fraction $\chi_{CO_{2}} \approx \frac{n_{CO_{2}}}{n_{H_{2}O}}$$n_{H_{2}O} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$$0.835 = 1670 	imes \frac{n_{CO_{2}}}{55.55}$$n_{CO_{2}} = \frac{0.835 	imes 55.55}{1670} = 0.02777 \ mol$Mass of $CO_{2} = n_{CO_{2}} 	imes 	ext{Molar mass of } CO_{2} = 0.02777 	imes 44 = 1.222 \ g$$X = 1.222 	imes 10^{3} 	imes 10^{-3} \ g = 1222 	imes 10^{-3} \ g$Thus,$X = 1222$.

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