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CET Karnataka Medical CET - Karnataka Medical 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) \[6N\]            

    B) \[2N\]

    C) \[4N\]            

    D) \[8N\]

    Correct Answer: B

    Solution :

    \[6.023\times {{10}^{23}}\] molecules of \[HCl=1\]mole \[HCl\]      Hence, \[1.2046\times {{10}^{24}}\]molecules of             \[HCl=\frac{1.2046\times {{10}^{24}}\times 1}{6.023\times {{10}^{23}}}\cong 2\] 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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