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BVP Medical BVP Medical Solved Paper-2002

  • question_answer
    Weight of \[KMn{{O}_{4}}\] required to prepare 250 ml of its \[\frac{N}{10}\]. solution if equivalent weight of  \[KMn{{O}_{4}}\]is 31.6, will be:

    A)  \[0.64g\]            

    B)  \[1.24g\]

    C)  \[1.78g\]            

    D)  \[0.79g\]

    Correct Answer: D

    Solution :

                    Milli equivalents of \[KMn{{O}_{4}}\] solution \[=\frac{1}{10}\times 250=25\] Equivalents of \[KMn{{O}_{4}}=\frac{25}{1000}=0.025\] Thus, required weight = number of equivalents \[\times \] equivalent weight.                                 \[=0.025\times 31.6\]                                 \[=0.79g\]          


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