\(6 \mathrm{~g}\) of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_8\right)\) and anthracene…
\(6 \mathrm{~g}\) 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 \(5.1 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) )
\(2.60,3.40\)
\(3.40,2.60\)
\(2.90,3.10\)
\(3.10,2.90\)
Solution
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.