A) \[-6.54{}^\circ C\]
B) \[6.54{}^\circ C\]
C) \[0.654{}^\circ C\]
D) \[-\text{ }0.654{}^\circ C\]
Correct Answer: D
Solution :
| [d] \[\because \Delta {{T}_{f}}={{k}_{f}}\times Molality\,of\,solution\] |
| \[\Delta {{T}_{b}}={{k}_{b}}\times Molality\,of\,solution\] |
| \[or\,\,\frac{\Delta {{T}_{f}}}{\Delta {{T}_{b}}}=\frac{{{k}_{f}}}{{{k}_{b}}}\] |
| Given that |
| \[\Delta \]Tb = T2 - T1 = 100.18 - 100 = 0.18 |
| kf for water = 1.86K kg mol-1 |
| kb for water = 0.512K kg mol-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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