Question

The molarity of a solution of ethanol in water in which the mole fraction of $C_2{H}_{5}OH$ is $0.040$ is ?

• A. $2.09$ M
• B. $2.31$ M
• C. $20.9$ M
• D. $23.1$ M

Answer 1

The correct option is B $2.31$ M

Solution:- (B) $2.31M$

Given that mole fraction of ethanol is $0.040$

As we know that

$\text{mole fraction}=\frac{\text{no. of moles}}{\text{total no. of moles}}$

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

$\Rightarrow \frac{n_{C_2{H}_{5}OH}}{n_{C_2{H}_{5}OH}+n_{H_{2}O}}=0.04.....\left(1\right)$

Since the solution is dilute, therefore water is approx. $1L$, i.e., $1000mL$

Let density of water to be $1g/ml$,

$\therefore \text{Weight of ! L of water}=\text{Volume}\times \text{density}=1000g$

Molecular weight of water $=18g$

$\therefore$ No. of moles of water $\left(n_{H_{2}O}\right)=\frac{1000}{18}=55.55\text{mol}$

Now from ${eq}^n\left(1\right)$, we have

$\frac{n_{C_2{H}_{5}OH}}{n_{C_2{H}_{5}OH}+55.55}=0.04$

$\Rightarrow n_{C_2{H}_{5}OH}=0.04(n_{C_2{H}_{5}OH}+55.55)$

$\Rightarrow 0.96n_{C_2{H}_{5}OH}=2.22$

$\Rightarrow n_{C_2{H}_{5}OH}=\frac{2.22}{0.96}=2.31$

As $2.31$ moles of ethanol are present in $1L$ of water.

Hence the molarity of the solution will be $2.31M$.