A) \[~-\,6.54\,{}^\circ C\]
B) \[6.54{}^\circ \,C\]
C) \[0.654{}^\circ \,C\]
D) \[-\,0.654{}^\circ \,C\]
Correct Answer: D
Solution :
\[\because \]\[\Delta {{T}_{f}}={{k}_{f}}\times \] Molality of solution \[\Delta {{T}_{b}}={{k}_{b}}\times \] Molality of solution or \[\frac{\Delta {{T}_{f}}}{\Delta {{T}_{b}}}=\frac{{{k}_{f}}}{{{k}_{b}}}\] Given that \[\Delta {{T}_{b}}={{T}_{2}}-{{T}_{1}}=100.18-100=0.18\] \[{{k}_{f}}\]for water\[=1.86\,K\,kg\,mo{{l}^{-1}}\] \[{{k}_{b}}\] for water\[=0.512\,K\,kg\,mo{{l}^{-1}}\] \[\therefore \] \[\frac{\Delta {{T}_{f}}}{0.18}=\frac{1.86}{0.512}\] \[\Delta {{T}_{f}}=\frac{1.86\times 0.18}{0.512}\] \[=0.6539\sim 0.654\] \[\Delta {{T}_{f}}={{T}_{1}}-{{T}_{2}}\] \[0.654=0{}^\circ C-{{T}_{2}}\] \[\therefore \] \[{{T}_{2}}=-\,0.654{}^\circ C\] (\[{{T}_{2}}\to \]Freezing point of aqueous urea solution)
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