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A camphor sample melts at 176^ 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^ C . The hydrocarbon is made up of 92.3 % of carbon. What is the molecular formula of the hydrocarbon?

Options

  1. AC ₆ H ₆
  2. BC ₁₂ H ₁₂
  3. CC ₄ H ₄
  4. DC ₈ H ₈

Correct answer

C. C ₄ H ₄

Step-by-step solution

Melting point of camphor =176^ 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^ C Thus, T_f for camphor =176-156.77=19.23^ C =19.23 ~K Depression in freezing point, aligned T_f & = K_f w_B M_B 1000 W_A 19.23 & = 40 0.02 1000 M_B 0.8 aligned Molar mass of solute (M_B )= 40 0.02 1000 19.23 0.8 =52 Empirical formula = CH Empirical formula weight =12+1=13 Multiple (n)= Molecular weight Empirical formula weight =

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