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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    Degree of dissociation of an acid \[HCl\] is 95%. 0.192 g of the acid is present in 0.5 L solution. The pH of the solution is approximately

    A)  3                                            

    B)  4

    C)  2                                            

    D)  1

    Correct Answer: C

    Solution :

    Number of moles  \[HCl=\frac{0.192}{36.5}=0.0053\] Volume = 0.5 L conc. of \[\text{HCl}=\frac{0.0053}{0.5}=1.06\times {{10}^{-2}}M\] \[[{{H}^{+}}]\] in \[HCl\] solution \[=1.06\times {{10}^{-2}}\times \frac{95}{100}\] \[=100.7\times {{10}^{-4}}=1\times {{10}^{-2}}\] \[pH=-\log \,[{{H}^{+}}]=2\]


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