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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    An aqueous solution containing 6.5 g of \[NaCl\]of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid \[NaOH\]. The volume of 1 M acetic acid required to neutralise \[NaOH\] obtained above is

    A)  \[1000\text{ }c{{m}^{3}}\]          

    B)  \[2000\text{ }c{{m}^{3}}\]

    C)  \[100\text{ }c{{m}^{3}}\]                            

    D)  \[200\text{ }c{{m}^{3}}\]

    Correct Answer: C

    Solution :

    Weight of pure \[NaCl=6.5\times 0.9=5.85\,g\] No. of equivalence of \[NaCl=\frac{5.85}{58.5}=0.1\] No. of equivalence of \[NaOH\] obtained = 0.1 Volume of 1 M acetic acid required for the neutralisation of \[NaOH=\frac{0.1\times 1000}{1}\]                                                 \[=100\text{ }c{{m}^{3}}\]


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