Question

$6g$ of a mixture of naphthalene $\left(C_{10}{H}_8\right)$ and anthracene $\left(C_{14}{H}_{10}\right)$ is dissolved in $300$ gram of benzene. If the depression in freezing point is $0.70K$, the composition of naphthalene and anthracene in the mixture respectively in $g$ are (molal depression constant of benzene is $5.1Kkgmo{l}^{-1}$ )

• A. 2.60, 3.40
• B. 3.40, 2.60
• C. 2.90, 3.10
• D. 3.10, 2.90

Answer 1

Answer: (B)

Given, weight of solvent $=300g$

Molality $=\frac{\text{ moles of solute }}{\text{ weight of solvent }(kg)}$

$K_f=5.1Kkg/mol$

Depression in freezing point,

$\Delta T_f=K_f\times m$

$0.70=5.1\times m$

$\frac{0.70}{5.1}=\frac{\text{ Total moles of solute }}{\text{ Total weight of solvent }(kg)}$

Lets, assume $xg$ napthalene is present,

$0.137=\frac{\frac{x}{w_{C_{10}{H}_8}}+\frac{6-x}{w_{C_{14}{H}_{10}}}\times 1000}{300}$

$0.041=\frac{x}{128g}+\frac{6-x}{178g}$

$\Rightarrow 25x=84.211$

$x$ ( napthalene ) $=3.4$

Now, anthracene $=6-x=6-3.4=2.6g$