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JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • question_answer
    Two elements A & B form compounds having molecular formulae \[A{{B}_{2}}\] and \[A{{B}_{4}}.\]When dissolved in 20.0 g of benzene 1.00 g of \[A{{B}_{2}}\] lowers f.p. by \[2.3{}^\circ C\] whereas 1.00 g of \[A{{B}_{4}}\]lowers f.p. by \[1.3{}^\circ C\]. The molal depression constant for benzene in 1000 g is 5.1. The atomic masses of A and B are

    A)  52, 48                  

    B)  42, 25

    C)  25, 42                  

    D)  None

    Correct Answer: C

    Solution :

     Let atomic masses of A and B be a and b amu respectively \[\therefore \]Molar mass of \[A{{B}_{2}}=(a+2b)g\,\text{mo}{{\text{l}}^{-1}}\] and Molar mass of \[A{{B}_{4}}=(a+4b)g\,\text{mo}{{\text{l}}^{-1}}\]For compound \[A{{B}_{2}}\] \[\Delta {{T}_{b}}={{K}_{b}}\times {{W}_{B}}\times 1000/{{W}_{A}}\times {{M}_{B}}\] \[2.3=5.1\times 1\times 1000/20.0\times (a+2b)\]?.I For compound \[A{{B}_{4}}\] \[1.3=5.1\times 1\times 1000/20.0\times (a+4b)\] ....II Solving (I) and (II),     a =25.49 b= 42.64


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