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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