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JEE Main & Advanced Chemistry Solutions / विलयन Question Bank Method of expressing concentration of solution

  • question_answer
    How much of NaOH is required to neutralise 1500 \[c{{m}^{3}}\] of 0.1 N  HCl  (At. wt. of  Na =23)                [KCET 2001]

    A)                 4 g          

    B)                 6 g

    C)                 40 g       

    D)                 60 g

    Correct Answer: B

    Solution :

           1500 \[c{{m}^{3}}\] of 0.1 N HCl have number of gm equivalence            \[=\frac{{{N}_{1}}\times {{V}_{1}}}{1000}=\frac{1500\times 0.1}{1000}=0.15\]                 \[\therefore 0.15\,gm.\] equivalent of NaOH \[=0.15\times 40=6\,gm.\]


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