Question

Mole fraction of ethanol in ethanol-water mixture is $0.25$. Hence, the percentage concentration of ethanol by weight of the mixture is:

• A. $25\%$
• B. $75\%$
• C. $46\%$
• D. $54\%$

Answer 1

The correct option is C $46\%$

${\chi }_2=\frac{n_2}{n_1+n_2}=0.25$
${\chi }_1=\frac{n_1}{n_1+n_2}=0.75$
$\frac{{\chi }_1}{{\chi }_2}=\frac{n_1}{n_2}=\frac{3}{1}\Rightarrow \frac{W_1\times M{w}_2}{M{w}_1\times W_2}=\frac{3}{1}$
$\frac{W_1\times 46}{18\times W_2}=3$
$\frac{W_1}{W_2}=1.17$
$\therefore \frac{W_2}{W_1+W_2}=0.46$

Therefore, % concentration of ethanol by weight in mixture $=46$%.

Hence, the correct option is $C$