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CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

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Question : 28 of 37
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SECTION -C
The freezing point of a solution containing 5 g5\ \mathrm{g} of benzoic acid (M=122 g mol−1)(M=122\ \mathrm{g}\ \mathrm{mol}^{-1}) in 35 g35\ \mathrm{g} of benzene is depressed by 2.94 K2.94\ \mathrm{K}. What is the percentage association of benzoic acid if it forms a dimer in solution?
(Kf  for benzene =4.9 K kg mol−1)(K_f\ \text{ for benzene } =4.9\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1})
Solution:  
Observed molar mass of benzoic acid:
  MB=  Kf×WBWA×ΔTf\;M_B=\;\frac{K_f \times W_B}{W_A \times \Delta T_f}
  WB=5 gm\;W_B=5\ \mathrm{gm}
  WA=0.035 kg\;W_A=0.035\ \mathrm{kg}
  MB=  4.9×50.035×2.94=  24.50.1029\;M_B=\;\frac{4.9 \times 5}{0.035 \times 2.94}=\;\frac{24.5}{0.1029}
  =238 g mol−1\;=238\ \mathrm{g}\ \mathrm{mol}^{-1}
  Kf=4.9 K kgmol−1\;K_f=4.9\ \mathrm{K}\ \mathrm{kgmol}^{-1}
  ΔTf=2.94\;\Delta T_f=2.94
     Normal molar mass of   C6H5COOH\;\;\text{ Normal molar mass of }\; \mathrm{C}_6\mathrm{H}_5\mathrm{COOH}
  =122 g mol−1\;=122\ \mathrm{g}\ \mathrm{mol}^{-1}
  i=     normal molar mass      observed molar mass   =  122238=0.513\;i=\;\frac{\;\text{ normal molar mass }\;}{\;\text{ observed molar mass }\;}=\;\frac{122}{238}=0.513
  %   of association of acid   (α)\;\% \;\text{ of association of acid }\;(\alpha)
  2 C6H5COOH⇌(C6H5COOH)2\;2\ \mathrm{C}_6\mathrm{H}_5\mathrm{COOH} \rightleftharpoons (\mathrm{C}_6\mathrm{H}_5\mathrm{COOH})_2
  n=2α=  i−11n−1=  0.513−112−1\;n=2 \alpha=\;\frac{i-1}{\frac{1}{n-1}}=\;\frac{0.513-1}{\frac{1}{2}-1} =  (−0.487)(−0.5)=0.974=\;\frac{(-0.487)}{(-0.5)}=0.974
Percentage association of acid =0.974×100==0.974 \times 100= 97.4%97.4 \%
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