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J & K CET Engineering J and K - CET Engineering Solved Paper-2015

  • question_answer
    0.5 molal solution of a solute in benzene shows a depression in freezing point equal to 2 K. Molal depression constant for benzene is \[\text{5}\,\text{K}\,\text{kg}\,\text{mo}{{\text{l}}^{-1}}.\]If the solute forms dimer in benzene, what is the % association?

    A)  40    

    B)  50    

    C)  60    

    D)  80

    Correct Answer: A

    Solution :

     Given, molality of solution, \[m=0.5\,m\] \[\Delta {{T}_{f}}=2K\] \[{{K}_{f}}=5K\,kg\,mo{{l}^{-1}}\] \[%\]of association, \[\alpha =?\] Step I Calculation of van?t Hoff factor (i) \[\Delta {{T}_{f}}=i{{K}_{f}}M\] \[i=\frac{\Delta {{T}_{f}}}{{{K}_{f}}M}\] \[i=\frac{2K}{5\,K\,kg\,mo{{l}^{-1}}\times 0.5\,m}\] \[i=\frac{4}{5}\] Step II Calculation of % association of acid \[(\alpha ).\] As solute form dimer in benzene, i.e.          \[2A{{A}_{2}}\] Therefore, value of \[x=2\] \[\alpha =\frac{i-1}{1/x-1}\] \[\alpha =\frac{4/5-1}{1/2-1}=0.4\] \[\therefore \] % of \[\alpha =0.4\times 100=40%\]


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