A camphor sample melts at 176^{circ} C. K_f f | vedclass

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(C) Melting point of pure camphor $= 176^{\circ} C$$K_f$ for camphor $= 40 \ K \ kg \ mol^{-1}$Mass of hydrocarbon solute $(w_B) = 0.02 \ g$Mass of ca

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$C_4 H_4$

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(C) Melting point of pure camphor $= 176^{\circ} C$$K_f$ for camphor $= 40 \ K \ kg \ mol^{-1}$Mass of hydrocarbon solute $(w_B) = 0.02 \ g$Mass of camphor solvent $(w_A) = 0.8 \ g$Freezing point of solution $= 156.77^{\circ} C$Depression in freezing point, $\Delta T_f = 176 - 156.77 = 19.23 \ K$Using the formula $\Delta T_f = \frac{K_f 	imes w_B 	imes 1000}{M_B 	imes w_A}$:$19.23 = \frac{40 	imes 0.02 	imes 1000}{M_B 	imes 0.8}$$M_B = \frac{40 	imes 0.02 	imes 1000}{19.23 	imes 0.8} = \frac{800}{15.384} \approx 52 \ g \ mol^{-1}$For the hydrocarbon with $92.3 \%$ carbon, the remaining $7.7 \%$ is hydrogen.Empirical formula calculation:$C: \frac{92.3}{12} = 7.69 \approx 7.7$$H: \frac{7.7}{1} = 7.7$Ratio $C:H = 1:1$, so empirical formula is $CH$ (empirical mass $= 13 \ g \ mol^{-1}$).$n = \frac{	ext{Molar mass}}{	ext{Empirical mass}} = \frac{52}{13} = 4$Molecular formula $= (CH)_4 = C_4 H_4$.

$A$ camphor sample melts at $176^{\circ} C$. $K_f$ for camphor is $40 \ K \ kg \ mol^{-1}$. $A$ solution of $0.02 \ g$ of a hydrocarbon in $0.8 \ g$ of camphor melts at $156.77^{\circ} C$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?

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