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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    Molecules of benzoic acid \[({{C}_{6}}{{H}_{5}}COOH)\]dimerise in benzene. V g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is (Given that \[{{K}_{f}}=5\,K\,kg\,mo{{l}^{-1}},\]Molar mass of benzoic acid \[=122g\,mo{{l}^{-1}}\])                   [JEE Main Online Paper Held On 12-Jan-2019 Evening]          

    A) 2.4 g                           

    B) 1.8 g

    C) 1.0 g               

    D)   1.5 g

    Correct Answer: A

    Solution :

    \[\Delta {{T}_{f}}=i{{k}_{f}}.m\]                      ?(i) where, m = molality \[\Delta {{T}_{f}}=\]depression in freezing point i = van?t Hoff factor and \[m=\frac{w}{122}\times \frac{1000}{30}\] For association,\[i=1+\left( \frac{1}{2}-1 \right)0.8=0.6\] So, form eqn. (i), \[2=0.6\times 5\times \frac{w}{122}\times \frac{1000}{30}\] or\[w=\frac{122\times 2\times 30}{0.6\times 5\times 100}=2.44g\]


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