A) 271K
B) 273.15K
C) 269.07K
D) 277.23K
Correct Answer: C
Solution :
[c] 269.07 K Depression in freezing point. \[\Delta {{T}_{f}}=\frac{1000\times kf\times {{w}_{2}}}{{{w}_{1}}\times {{M}_{2}}}\] \[\Delta {{T}_{f}}=(cane\,sugar\,)=273.15-271-{{2.15}^{{}^\circ }}\] \[\therefore 2.15=\frac{1000\times kf\times 5}{95\times 342}(1)\] \[\Delta {{T}_{f}}=(glucose\,)=\frac{1000\times kf\times 5}{95\times 180}(2)\] Dividing (2) by (1) \[\frac{\Delta {{T}_{f}}(glucose)}{2.15}=\frac{342}{180}\] \[Or\,\,\Delta {{T}_{f}}\,(glocose)=\frac{342}{180}\times \frac{2.15}{100}={{4.08}^{{}^\circ }}\] \[\therefore \,\] Freezing point of glucose solution = 273.15 - 4.08 = 269.07K
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