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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    5 g of\[N{{a}_{2}}S{{O}_{4}}\] was dissolved in x g of \[{{H}_{2}}O.\] The change in freezing point was found to be \[3.82{}^\circ C\]. If \[N{{a}_{2}}S{{O}_{4}}\] is 81.5% ionised, the value of x (\[{{K}_{f}}\] for \[water=1.86{}^\circ C\] kg \[\text{mo}{{\text{l}}^{\text{-1}}}\]) is approximately: (molar mass of S = 32 g \[\text{mo}{{\text{l}}^{\text{-1}}}\] and that of Na = 23 g\[\text{mo}{{\text{l}}^{\text{-1}}}\])            [JEE Online 08-04-2017]

    A)  25 g                                      

    B)  65 g

    C)  15 g                                      

    D)  45 g

    Correct Answer: D

    Solution :

    \[N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+SO_{4}^{2-}\] \[x=1+(3-1)0.815=2.63\] \[3.82=1.86\times 2.63\times \frac{5\times 1000}{142\times X}\] \[\therefore \]\[X=\frac{1.86\times 2.63\times 5000}{142\times 3.82}\]=45gm


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