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NEET Sample Paper NEET Sample Test Paper-52

  • question_answer
    0.116 g of \[{{C}_{4}}{{H}_{4}}{{O}_{4}}(A)\] is neutralised by 0.074 g of\[Ca{{\left( OH \right)}_{2}}\]. Hence protonic hydrogen \[({{H}^{+}})\] in will be-

    A) 1                                 

    B) 2

    C) 3                                 

    D) 4      

    Correct Answer: B

    Solution :

    Let the basicity of be x Then Eq. of \[\left( A \right)\,\,=\,\,Eq.\text{ }of\text{ }Ca{{\left( OH \right)}_{2}}\] \[\Rightarrow \,\,\,\,\frac{0.116}{\frac{116}{x}}\,\,=\,\,\frac{0.074}{\frac{74}{2}}\,\,\Rightarrow \,\,x\,\,=\,\,2\]


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