Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is $0.040$ (assume the density of water to be one).

Given,
Mole fraction of ethanol$=0.040$
$d_{H_{2}O}=1$
Step:-1 Calculation of mole of $C_2{H}_{5}OH$

We know, mole fraction of component in solution $=\frac{\text{moles of a particular component}}{\text{Total number of moles of solution}}$

So, Mole fraction of $C_2{H}_{5}OH=\frac{\text{Moles of}{C}_2{H}_{5}OH}{\text{Number of total moles of solution}}$

$0.040=\frac{n_{C_2{H}_{5}OH}}{n_{C_2{H}_{5}OH}+n_{H_{2}O}}$

To find molarity we have to find the number of moles of ethanol present in $1L$ of solution.

Calculation of moles of water present.
Now, number of moles present in $1L$ water.
$n_{H_{2}O}=\frac{1000g}{18gmo{l}^{-1}}$
$55.55mol$
Substitute this value in equation $1$

$0.040=\frac{n_{C_2{H}_{5}OH}}{n_{C_2{H}_{5}OH}+55.55}$

$n_{C_2{H}_{5}OH}=2.314mol$

Calculation of molarity

Now, molarity of $C_2{H}_{5}OH=\frac{n_{C_2{H}_{5}OH}}{\text{Volume of solution(inL)}}=\frac{2.314mol}{1L}$

Final answer: The molarity of the solution will be $2.314M$.