Question

The mole fraction of a given sample of $I_2$ in $C_6{H}_6$ is $0.2$. The molality of $I_2$ in $C_6{H}_6$ is

• A. $0.32$
• B. $3.2$
• C. $0.032$
• D. $0.48$

Answer 1

Answer: (B)

Mole fraction of $I_2=X_A$
$X_A=\frac{n_1}{n_1+n_2}$
$X_B=\frac{n_2}{n_1+n_2}$
$X_A+X_B=1$ $X_A=0.2$
$X_B=0.8$ $\frac{X_A}{X_B}=\frac{n_1}{n_2}$
$n_1:n_2=1:4$

Molality $=\frac{n_1}{\text{mass of solvent (in kg)}}$
$=\frac{1}{4\times 78}\times 1000=3.21m$