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Punjab Medical Punjab - MET Solved Paper-2004

  • question_answer
    \[Ca{{(OH)}_{2}}+{{H}_{3}}P{{O}_{4}}\xrightarrow{{}}CaHP{{O}_{4}}+2{{H}_{2}}O\]the equivalent weight of\[{{H}_{3}}P{{O}_{4}}\]in the above reaction is:

    A) \[21\]                                   

    B) \[27\]

    C) \[38\]                                   

    D) \[49\]

    Correct Answer: D

    Solution :

    The equivalent weight of\[{{H}_{3}}P{{O}_{4}}=\frac{molecular\,\,weight}{2}\](molecular wt. of\[{{H}_{3}}P{{O}_{4}}\]\[=3+31+64=98\])         \[=\frac{98}{2}=49\]


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