Question

$6 \,g$ 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 $0.70 \,K$, the composition of naphthalene and anthracene in the mixture respectively in $g$ are (molal depression constant of benzene is $5.1 \,K \,kg\, mol ^{-1}$ )

• 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 $=300\, g$

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

$K_{f}=5.1 \,K \,kg / mol$

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 $x\, g$ napthalene is present,

$0.137=\frac{\frac{x}{w_{ C _{10} H _{8}}}+\frac{6-x}{w_{ C _{14} H _{10}}} \times 1000}{300}$

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

$ \Rightarrow 25 x=84.211 $

$x$ ( napthalene ) $=3.4$

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