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J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    The van't Hoff factor for \[\text{0}\text{.1}\,\text{M}\,\text{Ba(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\] solution is 2.74. The degree of dissociation is

    A)  91.3%       

    B)  87%

    C)  100%        

    D)  74%

    Correct Answer: B

    Solution :

     \[Ba{{(N{{O}_{3}})}_{2}}B{{a}^{2+}}+2N{{O}_{3}}^{-}\] \[\begin{align}   & \text{At}\,\text{t}\,\text{= 0}\,\,\,\,\,\,\,0.1\,M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \\  & \text{At}\,\text{equ}\text{.}\,\,\,\,\,(0.1-x)M\,\,\,\,\,\,x\,M\,\,\,\,\,2x\,M \\ \end{align}\] \[i=\frac{(0.1-x)+x+2x}{0.1}\] \[2.74=\frac{0.1+2x}{0.1}\] \[2.74=\frac{0.1+2x}{0.1}\] \[0.1+2x=0.274\] \[2x=0.274-0.1=0.174\] \[x=\frac{0.174}{2}=0.087\] \[\therefore \]Degree of dissociation \[=\frac{0.087}{0.1}\times 100=87%\]


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