A) 273.07 K
B) 269.07 K
C) 273.15 K
D) 260.09 K
Correct Answer: B
Solution :
\[\Delta {{T}_{f}}=\frac{{{K}_{f}}\times {{W}_{B}}}{{{M}_{B}}\times {{W}_{A}}}\] For cane sugar solution, 2.15 K=\[\frac{{{K}_{f}}\times 5}{342\times 0.095}\] (95 g of water = 0.095 kg) For glucose solution, \[\Delta {{T}_{f}}=\frac{{{K}_{f}}\times 5}{180\times 0.095}\] \[\frac{\Delta {{T}_{f}}}{2.15}=\frac{{{K}_{f}}\times 5}{180\times 0.095}\times \frac{342\times 0.095}{{{K}_{f}}\times 5}\] \[\Delta {{T}_{f}}=\frac{342}{180}\times 2.15=4.085K\] Freezing point of glucose solution = 273.15 - 4.085 = 269.07 K
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