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

Question

$6 \mathrm{~g}$ of a mixture of naphthalene $\left(\mathrm{C}_{10} \mathrm{H}_8\right)$ and anthracene $\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)$ is dissolved in 300 gram of benzene. If the depression in freezing point is $0.70 \mathrm{~K}$, the composition of naphthalene and anthracene in the mixture respectively in $\mathrm{g}$ are (molal depression constant of benzene is $5.1 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ), $2.60,3.40$, $3.40,2.60$, $2.90,3.10$, $3.10,2.90$

MARKS App · Accepted Answer

Given, weight of solvent $=300 \mathrm{~g}$ $\begin{aligned} \text {Molality } & =\frac{\text { moles of solute }}{\text { weight of solvent }(\mathrm{kg})} \ K_f & =5.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol} \end{aligned}$ Depression in freezing point, $\begin{gathered} \Delta T_f=K_f \times m \ 0.70=51 \times m \ \frac{0.70}{5.1}=\frac{\text { Total moles of solute }}{\text { Total weight of solvent }(\mathrm{kg})} \end{gathered}$ Lets, assume $x$ g napthalene is present, $\begin{aligned} 0.137 & =\frac{\frac{x}{w_{\mathrm{C}_{10} \mathrm{H}_8}}+\frac{6-x}{w_{\mathrm{C}_{14} \mathrm{H}_{10}}} \times 1000}{300} \ 0.041 & =\frac{x}{128 \mathrm{~g}}+\frac{6-x}{178 \mathrm{~g}} \Rightarrow 25 x=84.211 \ x(\text {napthalene}) & =3.4 \ \text {Now, anthracene } & =6-x=6-3.4=2.6 \mathrm{~g} \end{aligned}$ Hence, option (2) is correct.

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