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J & K CET Medical J & K - CET Medical Solved Paper-2003

  • question_answer
    An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N sodium hydroxide required to completely neutralise 10 mL of this solution is:

    A)  40 mL                                  

    B)  20 mL

    C)  10 mL                                  

    D)  4mL

    Correct Answer: A

    Solution :

                    \[W=\frac{NEV}{1000}\] \[N=\frac{W\times 1000}{E\times V}=\frac{6.3\times 1000}{63\times 250}=0.4\,N\] \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[0.1\times {{V}_{1}}=0.4\times 10\] \[{{V}_{1}}=\frac{0.4\times 10}{0.1}\] \[{{V}_{1}}=40\,mL\]


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