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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    Two elements A & B form compounds having molecular formulae \[A{{B}_{2}}\] and \[A{{B}_{4}}\]. When dissolved in \[20.0\text{ }g\] of benzene \[1.00g\] of \[A{{B}_{2}}\] lowers f.p. by \[{{2.3}^{o}}C\] whereas \[1.00g\] of \[A{{B}_{4}}\]lowers f.p. by\[{{1.3}^{o}}C\]. The molal depression constant for benzene in \[1000g\] 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 \[A{{B}_{2}}=(a+2b)g\,mo{{l}^{-1}}\]and Molar mass of \[A{{B}_{4}}=(a+4b)g\,mo{{l}^{-1}}\] 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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