Calculate the molarity of ethanol in a soluti | vedclass

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(A) Molarity is defined as the number of moles of solute per liter of solution.Assuming $1 \ L$ of water,the number of moles of $H_2O$ is $n_{H_2O} =

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$2.31$

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(A) Molarity is defined as the number of moles of solute per liter of solution.Assuming $1 \ L$ of water,the number of moles of $H_2O$ is $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.For a binary solution,the sum of mole fractions is $X_{H_2O} + X_{C_2H_5OH} = 1$.Given $X_{C_2H_5OH} = 0.040$,then $X_{H_2O} = 1 - 0.040 = 0.96$.The mole fraction formula is $X_{H_2O} = \frac{n_{H_2O}}{n_{H_2O} + n_{C_2H_5OH}}$.Substituting the values: $0.96 = \frac{55.55}{55.55 + n_{C_2H_5OH}}$.Solving for $n_{C_2H_5OH}$: $53.328 + 0.96 	imes n_{C_2H_5OH} = 55.55$.$0.96 	imes n_{C_2H_5OH} = 2.222$.$n_{C_2H_5OH} = \frac{2.222}{0.96} \approx 2.3145 \ mol$.Since this is in $1 \ L$ of solution,the molarity is $2.31 \ M$.

Calculate the molarity of ethanol in a solution where the mole fraction of ethanol is $0.040$. (Assume the density of water is $1 \ g/mL$.) (in $M$)

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