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JIPMER Jipmer Medical Solved Paper-2015

  • question_answer
    2.5 g of the carbonate of a metal was treated with 100 ml of\[1\,N\]\[{{H}_{2}}S{{O}_{4}}\]. After the completion of the reaction, the solution was boiled off to expel \[C{{O}_{2}}\] and was then titrated against \[1\,N\]\[NaOH\]solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20.

    A)  50  

    B)     25                         

    C)     75                                         

    D)     100

    Correct Answer: A

    Solution :

      Equivalent weight of metal carbonate = 20 + 30 = 50 2.5 g of metal carbonate\[=\frac{2.5}{50}=0.05\,\,\text{eq}\text{.}\] Number of equivalent of \[{{H}_{2}}S{{O}_{4}}\] would have reacted = 0.05. Number of equivalent of \[{{H}_{2}}S{{O}_{4}}\]taken \[=\frac{100\times 1}{1000}=0.1\] Number of equivalent of\[{{H}_{2}}S{{O}_{4}}\]remains unreacted = 0.1 - 0.05 = 0.05 eq. \[\therefore \] Number of equivalent of alkali consumed = 0.05 eq. milli eq. = Normality\[\times \]Volume in mL. \[\therefore \]\[1.0\times V=0.05\times 1000\] \[V=\frac{0.05\times 1000}{1.0}=50\,mL\]                           


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