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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2014

  • question_answer
    A solution of \[1.25g\]of P in \[50g\]of water lowers freezing point by \[{{0.3}^{o}}C\]. Molar mass of P is \[{{K}_{f(water)}}=1.86K\,kg\,mo{{l}^{-1}}\]. The degree  of association of Pin water is              

    A)  \[80%\]

    B)  \[60%\]

    C)   \[65%\]

    D)  \[75%\]

    Correct Answer: A

    Solution :

    : Given: \[{{K}_{f}}=1.86K\text{ }kg\text{ }mo{{l}^{-1}},w=1.25\text{ }g,\] \[~W=50g,Ar={{0.3}^{o}}C,M=?\] As P undergoes association, \[2P{{(P)}_{2}};n=2\] Using equation,     \[M=\frac{1000\times {{K}_{f}}\times w}{W\times \Delta T}\]     \[M=\frac{1000\times 1.86\times 1.25}{50\times 0.3}=155\] Now, \[i=\frac{Normal\,mol.\,mass}{Observed\,mol.\,mass}=\frac{94}{155}=0.606\] Then, the degree of association of P is     \[\alpha =\frac{1-i}{1-\frac{1}{n}}\] \[\alpha =\frac{1-0.606}{1-\frac{1}{2}}=0.788\] or \[78.8%\sim 80%\]


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