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BVP Medical BVP Medical Solved Paper-2013

  • question_answer
    An aqueous solution containing 1.248 g of barium     chloride     (molar    mass \[=208.34g\,mo{{l}^{-1}}\]) in 100 g of water boils at\[{{100.0832}^{o}}C\]. The degree of dissociation of \[BaC{{l}_{2}}\] will be (\[{{K}_{b}}\] for water\[=0.52K\text{ }kg\text{ }mo{{l}^{-1}}\])

    A)  \[0.835\]                            

    B)  \[0.756\]

    C)  \[0.638\]                            

    D)  \[0.930\]

    Correct Answer: A

    Solution :

                    \[{{M}_{2}}\](observe\[=\frac{1000\times 0.52\times 1.248}{100\times 0.0832}=78g\,\,mo{{l}^{-1}}\] \[i=\frac{208.34}{78}=2.67\]                                 \[\underset{1-\alpha }{\mathop{\underset{1\,mol}{\mathop{BaC{{l}_{2}}}}\,}}\,\xrightarrow{{}}\underset{\alpha }{\mathop{\underset{0}{\mathop{B{{a}^{2+}}}}\,}}\,+\underset{\begin{smallmatrix}  0 \\  2\alpha  \end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\,\] Number of total moles \[=1+2\alpha \] or \[i=1+2\alpha \] or \[\alpha =\frac{i-1}{2}=\frac{2.67-1}{2}=0.835\]


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