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BCECE Engineering BCECE Engineering Solved Paper-2004

  • question_answer
    The normality of mixture obtained by mixing 100 mL of \[\text{0}\text{.2 M }{{\text{H}}_{\text{2}}}\text{S}{{\text{0}}_{\text{4}}}\text{+ 100 mL}\]of \[\text{0}\text{.2 M NaOH}\]is:

    A) 0.2                         

    B)        0.01

    C) 0.1                         

    D)        0.3

    Correct Answer: C

    Solution :

    Given \[{{H}_{2}}S{{O}_{4}}-V=100\,mL\,\,\,N=0.2\,M\] \[NaOH-V=100\,mL\,\,N=0.2\,M\]           Milliequivalent of \[{{H}_{2}}S{{O}_{4}}=100\times 0.2\times 2=40\] (\[\therefore \] it is dibasic acid) Milliequivalent of \[NaOH=100\times 0.1\times 1=20\] \[\therefore \] Milliequivalent of \[{{H}_{2}}S{{O}_{4}}\]left \[~=40-20=20\] Total volume \[=100\text{ }mt+100\text{ }mL=200\text{ }mL\] Normality of \[{{H}_{2}}S{{O}_{4}}(left)=\frac{20}{200}=0.1\,N\]


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