A) -6.54°C
B) 6.54°C
C) 0.654°C
D) - 0.654°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 \[=1.86K\,kg\,mo{{l}^{-1}}\] \[{{k}_{b}}\] for water \[=0.512K\,\log \,mo{{l}^{-1}}\] \[\therefore \frac{\Delta {{T}_{f}}}{0.18}=\frac{1.86}{0.512}\] \[\Delta {{T}_{f}}=\frac{1.86\times 0.18}{0.512}\] \[=0.6539\tilde{\ }0.654\] \[\Delta {{T}_{f}}={{T}_{1}}-{{T}_{2}}\] \[0.654=0{}^\circ C-{{T}_{2}}\] \[\therefore {{T}_{2}}=-0.654{}^\circ C\] (\[{{T}_{2}}\to \] Freezing point of aqueous urea solution)
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