A) \[0.654\,{}^\circ C\]
B) \[-\,0.654\,{}^\circ C\]
C) \[\,6.54\,{}^\circ C\]
D) \[-\,6.54\,{}^\circ C\]
Correct Answer: B
Solution :
As \[\Delta {{T}_{f}}={{K}_{f}}\,m\] \[\Delta {{T}_{b}}=\,\,{{K}_{b}}.\,m\] Hence, we have \[m=\frac{\Delta {{T}_{f}}}{{{K}_{f}}}\,\,=\,\,\frac{\Delta {{T}_{b}}}{{{K}_{b}}}\] \[\Delta {{T}_{f}}=\Delta {{T}_{b}}\frac{{{K}_{f}}}{{{K}_{b}}}\] \[\Rightarrow \,\,\,[\Delta {{T}_{b}}=100.18-100=0.18{}^\circ \,C]\] \[=\,\,0.18\times \frac{1.86}{0.512}=0.654{}^\circ \,C\] As the Freezing Point of pure water is \[0{}^\circ C\], \[\Delta {{T}_{f}}=0-{{T}_{f}}\] \[0.654=\,\,0-{{T}_{f}}.\] \[\therefore \,\,{{T}_{f}}=-\,0.654\] Thus the freezing point of solution will be \[-\,0.654{}^\circ C\].
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