36 g of water and 828 g of ethyl alcohol fo | vedclass

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(D) Step $1$: Calculate the number of moles of water $(n_{H_2O})$.$n_{H_2O} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{36 \ g}{18 \ g/mol} = 2 \

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$0.1$

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(D) Step $1$: Calculate the number of moles of water $(n_{H_2O})$.$n_{H_2O} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{36 \ g}{18 \ g/mol} = 2 \ mol$.Step $2$: Calculate the number of moles of ethyl alcohol $(n_{C_2H_5OH})$.$n_{C_2H_5OH} = \frac{828 \ g}{46 \ g/mol} = 18 \ mol$.Step $3$: Calculate the mole fraction of water $(x_{H_2O})$.$x_{H_2O} = \frac{n_{H_2O}}{n_{H_2O} + n_{C_2H_5OH}} = \frac{2}{2 + 18} = \frac{2}{20} = 0.1$.

$36 \ g$ of water and $828 \ g$ of ethyl alcohol form an ideal solution. The mole fraction of water in it is:

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