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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    How much of\[\text{NaOH}\]is required to neutralise \[\text{1500}\,\text{c}{{\text{m}}^{\text{3}}}\] of \[\text{0}\text{.1}\,\text{N}\,\text{HCl?}\]\[\text{(Na}\,\text{=}\,\text{23):}\]

    A)  60 g                                      

    B)  6 g

    C)  4 g                                        

    D)  40 g

    Correct Answer: B

    Solution :

    \[1500\,c{{m}^{3}}\]of \[0.\text{1}\,\text{N}\,\text{HCl}\]have number of gm equivalence \[=\frac{{{N}_{1}}\times {{V}_{1}}}{1000}=\frac{1500\times 0.1}{1000}=0.15\]                 \[\because \]     0.15 gm. equivalent of\[\text{NaOH}\] \[=1.5\times 40=6\,gm.\]            


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