Calculate the mole fraction of C_2H_6O_2 (eth | vedclass

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(B) Given: Mass percentage of $C_2H_6O_2 = 28 \%$. This means $28 \ g$ of $C_2H_6O_2$ is present in $100 \ g$ of the solution. Mass of water = $100 \

Vedclass · Accepted Answer

$0.101$

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(B) Given: Mass percentage of $C_2H_6O_2 = 28 \%$. This means $28 \ g$ of $C_2H_6O_2$ is present in $100 \ g$ of the solution. Mass of water = $100 \ g - 28 \ g = 72 \ g$. Molar mass of $C_2H_6O_2 = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g/mol$. Molar mass of $H_2O = 18 \ g/mol$. Moles of $C_2H_6O_2 (n_1) = \frac{28}{62} \approx 0.4516 \ mol$. Moles of $H_2O (n_2) = \frac{72}{18} = 4.000 \ mol$. Mole fraction of $C_2H_6O_2 (x_1) = \frac{n_1}{n_1 + n_2} = \frac{0.4516}{0.4516 + 4.000} = \frac{0.4516}{4.4516} \approx 0.1014$. Therefore,the mole fraction is approximately $0.101$.

Calculate the mole fraction of $C_2H_6O_2$ (ethylene glycol) in an aqueous solution containing $28 \%$ by mass of $C_2H_6O_2$.

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