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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    35.4 mL of \[\text{HCl}\] is required for the neutralization of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is?

    A)  0.97 N                                 

    B)  0.142 N

    C)  0.194N                                

    D)  0.244 N

    Correct Answer: C

    Solution :

                     We know that 1g equivalent weight of \[NaOH=40\,g\]                 \[\therefore \]\[40\,g\]of\[NaOH=1\,g\]eq. of \[NaOH\] \[\therefore \] \[0.275\,g\]\[NaOH=\frac{1}{40}\times 0.275\,eq.\]                                 \[=\frac{1}{40}\times 0.275\times 1000\]                                 \[=6.88\,\text{meq}\] \[\underset{(HCl)}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{(NaOH)}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[{{N}_{1}}\times 35.4=6.88\]    \[(\because \,meq=NV)\] \[{{N}_{1}}=0.194\]


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