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CMC Medical CMC-Medical Ludhiana Solved Paper-2013

  • question_answer
    Two elements A and B form compounds having molecular formula \[A{{B}_{2}}\] and \[A{{B}_{4}}.\] When dissolved in 20 g of benzene \[{{C}_{6}}{{H}_{6}},\]\[1\,g\] of \[A{{B}_{2}}\]lowers the freezing point by 2.3 K whereas 1.0 g of \[A{{B}_{4}}\]lowers it by 1.3 K. The molal depression constant for benzene is 5.1K kg \[\text{mo}{{\text{l}}^{-1}}\]. Calculate atomic masses of A and B.

    A)  A = 25.59 u, B = 42.64 u

    B)  A = 36.78 u, B = 49.38 u

    C)  A = 42.64 u, B = 25.59 u

    D)  A = 49.38 u, B = 36.78 u

    Correct Answer: A

    Solution :

                    \[{{M}_{2}}=\frac{1000\,{{K}_{f}}{{w}_{2}}}{{{w}_{1}}\times \Delta {{T}_{f}}}\] \[{{M}_{A{{B}_{2}}}}=\frac{1000\times 5.1\times 1}{20\times 2.3}=110.87\,g\,mo{{l}^{-1}}\] \[{{M}_{A{{B}_{4}}}}=\frac{1000\times 5.1\times 1}{20\times 1.3}=196.15\,g\,mo{{l}^{-1}}\] Suppose atomic masses of A and B are a and b respectively. Then Molar mass of \[A{{B}_{2}}=a+2b=110.87g\,mo{{l}^{-1}}\]            ...(i) Molar mass of \[A{{B}_{4}}=a+4b=196.15\,g\,mo{{l}^{-1}}\]         ...(ii) On subtracting Eq (i) from (ii) 2b = 85.28 or b = 42.64 On substituting value of b in Eq. (i) \[a+2\times 42.64=110.87\]                 \[a=25.59\] Atomic mass of A = 25.59 u Atomic mass of B = 42.64 u


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