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Q. The mole fraction of a given sample of $I_2$ in $C_6H_6$ is $0.2$. The molality of $I_2$ in $C_6H_6$ is

Some Basic Concepts of Chemistry

Solution:

Mole fraction of $I_2 = X_A$
$X_A = \frac{n_1}{n_1 + n_2}$
$X_B = \frac{n_2}{n_1 + n_2}$
$X_A + X_B = 1$ $X_A = 0.2$
$X_B = 0.8$ $\frac{X_A}{X_B} = \frac{n_1}{n_2}$
$n_1 : n_2 = 1 : 4$
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Molality $ = \frac{n_1}{\text{mass of solvent (in kg)}}$
$ = \frac{1}{4\times 78} \times 1000 = 3.21\,m$