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NEET Sample Paper NEET Sample Test Paper-17

  • question_answer
    Pure benzene freezes at \[5.3{}^\circ C\]. A solution of 0.223 g of phenylacetic acid \[({{C}_{6}}{{H}_{5}}C{{H}_{2}}COOH)\] in 4.4 g of benzene \[({{K}_{f}}=5.12\,K\,kg\,mo{{l}^{-1}})\] freezes at \[4.47{}^\circ C\]. From this observation, one can conclude that

    A)  Phenylacetic acid undergoes partial ionization in benzene.

    B)  Phenylacetic acid exists as such in benzene.

    C)  Phenylacetic acid dimerizes in benzene.

    D)  Phenylacetic acid undergoes complete ionization in benzene.

    Correct Answer: C

    Solution :

    \[\Delta {{\Tau }_{f}}={{T}_{0}}-{{T}_{f}}=5.3-4.47=0.83\] Molality of solvent \[m=\frac{0.223}{1.36}\times \frac{1000}{4.4}=0.373.\] As molecular weight of phenylacetic acid =136. \[{{K}_{f}}=5.12\] \[\Delta {{T}_{f}}=i.\,{{K}_{f}}.m\] So \[i=\frac{\Delta {{T}_{f}}}{{{K}_{f}}m}=\frac{0.83}{5.12\times 0.373}=0.45.\] As i is nearly equal to 0.5 it means phenylacetic acid undergoes dimerization in benzene. Hence, the correct option is (c).


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