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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    The normality of a solution containing \[31.5\text{ }gm\]. of hydrated oxalic acid \[({{C}_{2}}{{H}_{2}}{{O}_{4}}.2{{H}_{2}}O)\] in\[1250\text{ }ml\]. of solution is

    A)  \[0.1N\]                             

    B)         \[0.2N\]

    C)         \[0.4N\]                             

    D)         \[0.6N\]

    Correct Answer: C

    Solution :

    Wt. of oxalic acid \[=31.5\text{ }gm.\] Volume of solution \[=1250\text{ }ml.\] \[\because \] Equivalents of oxalic acid \[=\frac{31.5}{63}=0.5\] \[\therefore \] Normality of solution \[=\frac{0.5}{1250}\times 1000\]                                                                 \[=0.4N\]


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