Discussion Forum

A  camphor  sample melts at $176 ^{0} C.$ $K_{f}$ for  camphor is 40 K kg mol^-1 . A solution of 0.02g  of a hydrocarbon  in 0.8 g of camphor melts at $156.77^{0}C$ . Hydrocarbon is  made up of  92.3% of carbon. What is the molecular formula of the hydrocarbon?

Answer:

Explanation:

Melting point of camphor= $176 ^{0}$ C

$ (K_{f})$ for camphor=40 K kg mol^-1

 Mass of hydrocarbon (w_B)=0.02 g

Mass of camphor (w_A)=0.8 g

 Temperature of camphor (at which it melts)= $156.77^{0} C$

 Thus, $\triangle T_{f}$ for  camphor =176-156-77=   $19.23^{0} C$=19.23 K

 $\therefore$ depression in freezing point ,

     $\triangle T_{f}$ =  $\frac{k_{f}\times w_{B}}{M_{B}}\times\frac{1000}{w_{A}}$

              $19.23= \frac{40\times0.02\times1000}{M_{B}\times0.8}$

 Molar mass of solute (M_B)

=$\frac{40\times0.02\times1000}{19.23\times0.8}=52$

 $\therefore$   Empricial formula = CH

 Empirical  formula  weight =12+1=13

 Multiple (n) = Molecular weight/ Empirical  formula weight = $\frac{52}{13}=4$

 Thus, molecular formula = (CH)₄= C₄H₄