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Manipal Medical Manipal Medical Solved Paper-2000

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


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