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A camphor sample melts at $176^{\circ} \mathrm{C} . K_f$ for camphor is $40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. A solution of $0.02 \mathrm{~g}$ of a hydrocarbon in $0.8 \mathrm{~g}$ of camphor melts at $156.77^{\circ} \mathrm{C}$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?
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The correct answer is:
$\mathrm{C}_4 \mathrm{H}_4$
Melting point of camphor $=176^{\circ} \mathrm{C}$
$\left(K_f\right)$ for camphor $=40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Mass of hydrocarbon $\left(w_B\right)=0.02 \mathrm{~g}$
Mass of camphor $\left(w_A\right)=0.8 \mathrm{~g}$
Temperature of camphor (at which it melts)
$=156.77^{\circ} \mathrm{C}$
Thus, $\Delta T_f$ for camphor $=176-156.77$
$=19.23^{\circ} \mathrm{C}=19.23 \mathrm{~K}$
$\therefore$ Depression in freezing point,
$\begin{aligned} \Delta T_f & =\frac{K_f \times w_B}{M_B} \times \frac{1000}{W_A} \\ 19.23 & =\frac{40 \times 0.02 \times 1000}{M_B \times 0.8}\end{aligned}$
Molar mass of solute $\left(M_B\right)$
$=\frac{40 \times 0.02 \times 1000}{19.23 \times 0.8}=52$
$\therefore$ Empirical formula $=\mathrm{CH}$
Empirical formula weight $=12+1=13$
Multiple $(n)=\frac{\text { Molecular weight }}{\text { Empirical formula weight }}=\frac{52}{13}=4$
Thus, molecular formula $=(\mathrm{CH})_4=\mathrm{C}_4 \mathrm{H}_4$
$\left(K_f\right)$ for camphor $=40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Mass of hydrocarbon $\left(w_B\right)=0.02 \mathrm{~g}$
Mass of camphor $\left(w_A\right)=0.8 \mathrm{~g}$
Temperature of camphor (at which it melts)
$=156.77^{\circ} \mathrm{C}$
Thus, $\Delta T_f$ for camphor $=176-156.77$
$=19.23^{\circ} \mathrm{C}=19.23 \mathrm{~K}$
$\therefore$ Depression in freezing point,
$\begin{aligned} \Delta T_f & =\frac{K_f \times w_B}{M_B} \times \frac{1000}{W_A} \\ 19.23 & =\frac{40 \times 0.02 \times 1000}{M_B \times 0.8}\end{aligned}$
Molar mass of solute $\left(M_B\right)$
$=\frac{40 \times 0.02 \times 1000}{19.23 \times 0.8}=52$
$\therefore$ Empirical formula $=\mathrm{CH}$
Empirical formula weight $=12+1=13$
Multiple $(n)=\frac{\text { Molecular weight }}{\text { Empirical formula weight }}=\frac{52}{13}=4$
Thus, molecular formula $=(\mathrm{CH})_4=\mathrm{C}_4 \mathrm{H}_4$
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