A mixture has 18 g of water and 414 g of et | vedclass

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(A) $1$. Calculate the number of moles of ethanol $(C_2H_5OH)$: $n_{	ext{ethanol}} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{414 \ g}{46 \ g/mo

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$0.1$

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(A) $1$. Calculate the number of moles of ethanol $(C_2H_5OH)$: $n_{	ext{ethanol}} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{414 \ g}{46 \ g/mol} = 9 \ mol$. $2$. Calculate the number of moles of water $(H_2O)$: $n_{	ext{water}} = \frac{18 \ g}{18 \ g/mol} = 1 \ mol$. $3$. Calculate the mole fraction of water $(x_{	ext{water}})$: $x_{	ext{water}} = \frac{n_{	ext{water}}}{n_{	ext{water}} + n_{	ext{ethanol}}} = \frac{1}{1 + 9} = \frac{1}{10} = 0.1$.

$A$ mixture has $18 \ g$ of water and $414 \ g$ of ethanol. The mole fraction of water in the mixture is (assume ideal behavior of the mixture):

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