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JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    50 mL of 5.6% KOH (w/v) is added to 50 mL of a 5.6% \[HCl\] (w/v) solution. The resulting solution will be

    A)  neutral                         

    B)  alkaline

    C)  strongly alkaline            

    D)  acidic

    Correct Answer: D

    Solution :

    Milimoles of KOH \[=\frac{5.6\times 1000}{56\,\times 100}\,50=50\] Milimoles of \[HCl=\frac{5.6\times 1000}{36.5\,\times 100}\,\times 50\,=76.71\] So resulting solution will be acidic.


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