A mixture of alcohol and water contains 54 % | vedclass

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Question

(A) Given that the mixture contains $54 \%$ water by mass,we assume $100 \ g$ of the mixture.Mass of $H_2O = 54 \ g$.Mass of ethanol $(C_2H_5OH)$ $= 1

Vedclass · Accepted Answer

$0.25$

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(A) Given that the mixture contains $54 \%$ water by mass,we assume $100 \ g$ of the mixture.Mass of $H_2O = 54 \ g$.Mass of ethanol $(C_2H_5OH)$ $= 100 \ g - 54 \ g = 46 \ g$.Moles of $H_2O = \frac{54 \ g}{18 \ g/mol} = 3 \ mol$.Moles of ethanol $= \frac{46 \ g}{46 \ g/mol} = 1 \ mol$.Total moles $= 3 + 1 = 4 \ mol$.Mole fraction of ethanol $= \frac{	ext{moles of ethanol}}{	ext{total moles}} = \frac{1}{4} = 0.25$.

$A$ mixture of alcohol and water contains $54 \%$ water by mass. What is the mole fraction of ethanol?

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