Question

An X molal solution of a compound in benzene has mole fraction of solute = 0.2. The value of X is

• A. $14.0$
• B. $3.2$
• C. $1.4$
• D. $2.0$

Answer 1

Answer: (B)

Mole fraction of solute $X_2=0.2$ Therefore, mole fraction of solvent $X_1=0.8$
Or $\frac{n_2}{n_1+n_2}=0.2$ and $\frac{n_1}{n_1+n_2}=0.8$
or $\frac{n_2}{n_1}=\frac{0.2}{0.8}=\frac{1}{4}$
Now, if $n_1$ (solvent moles) $=1000/78=12.8$ moles
$n_2=12.8/4=3.2$ moles. Therefore, 3.2 moles of the compound are present in one kg of solvent benzene and so molality $=3.2$