138 g of ethyl alcohol is mixed with 72 g o | vedclass

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(A) Number of moles of ethyl alcohol $(C_2H_5OH)$ $= \frac{138 \ g}{46 \ g/mol} = 3 \ mol$. Number of moles of water $(H_2O)$ $= \frac{72 \ g}{18 \ g/

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$3: 4$

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(A) Number of moles of ethyl alcohol $(C_2H_5OH)$ $= \frac{138 \ g}{46 \ g/mol} = 3 \ mol$. Number of moles of water $(H_2O)$ $= \frac{72 \ g}{18 \ g/mol} = 4 \ mol$. Mole fraction of alcohol $(X_{C_2H_5OH})$ $= \frac{3}{3+4} = \frac{3}{7}$. Mole fraction of water $(X_{H_2O})$ $= \frac{4}{3+4} = \frac{4}{7}$. The ratio of mole fraction of alcohol to water is $\frac{X_{C_2H_5OH}}{X_{H_2O}} = \frac{3/7}{4/7} = \frac{3}{4}$.

$138 \ g$ of ethyl alcohol is mixed with $72 \ g$ of water. The ratio of mole fraction of alcohol to water is

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