JCECE Medical JCECE Medical Solved Paper-2010

  • question_answer What is the freezing point of a solution containing \[\text{8}\text{.1 g HBr}\]in 100 g water assuming the acid to be 90% ionised? (\[{{k}_{f}}\]for water\[=1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}}\])

    A) \[0.85{{\,}^{o}}C\]

    B)  \[-3.53{{\,}^{o}}C\]

    C)  \[0{{\,}^{o}}C\]

    D)  \[-0.35{{\,}^{o}}C\]

    Correct Answer: B

    Solution :

    \[\Delta {{\Tau }_{f}}=i\times {{k}_{f}}\times m\] \[\text{Ions}\,\text{at}\,\text{equilibrium}\overset{HBr}{\mathop{1-\alpha }}\,\xrightarrow{{}}\underset{\alpha }{\mathop{{{H}^{+}}}}\,+\underset{\alpha }{\mathop{B{{r}^{-}}}}\,\] \[\therefore \] \[\text{Total}\,\text{ions}=1-\alpha +\alpha +\alpha \] \[=1+\alpha \] \[\therefore \] \[i=1+\alpha \] Given, \[{{k}_{f}}=1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}}\] mass of\[\text{ }\!\!~\!\!\text{ HBr =}\,\text{8}\text{.1 g}\] mass of\[\text{ }\!\!~\!\!\text{ }{{\text{H}}_{\text{2}}}\text{O}\,\text{=100 g}\] \[\text{( }\!\!\alpha\!\!\text{ )}\,\,\text{=}\]degree of ionization \[=90%\] \[m\](molality) \[\text{= }\frac{\text{mass}\,\text{of}\,\text{solute/mol}\text{.wt}\text{.of}\,\text{solute}}{\text{mass of solventin kg}}\] \[=\frac{8.1/81}{100/1000}\] \[i=1+\alpha \] \[=1+90/100\] \[=1.9\] \[\Delta {{T}_{f}}=i\times {{k}_{f}}\times m\] \[=1.9\times 1.86\times \frac{8.1/81}{100/1000}\] \[=3.534{{\,}^{o}}C\] \[\Delta {{T}_{f}}=\]  (depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0-freezing point of solution \[\therefore \]Freezing point of solution\[=-3.534{{\,}^{o}}C\]

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