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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    A solution contains \[1.2046\times {{10}^{-24}}\] hydrochloric acid molecules in one \[d{{m}^{3}}\] of the solution. The strength of the solution is :

    A)  6 N                                       

    B)  2 N

    C)  4 N                                       

    D)  8 N

    Correct Answer: B

    Solution :

    \[6.023\times {{10}^{23}}\] molecules of \[HCl\equiv 1\] mole                                                                 \[HCl\] Hence, \[1.2046\times {{10}^{24}}\] molecules of                 \[HCl\equiv \frac{1.2046\times {{10}^{24}}\times 1}{6.023\times {{10}^{23}}}\equiv 2\] mole \[HCl\] Thus two moles (= two gram - equivalents) of \[HCl\] are dissolved in one \[d{{m}^{3}}\](-one litre) solution. Therefore the solution will be 2N.


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