A) 272.898 K
B) \[0.102{}^\circ C\]
C) 273 K
D) \[0.108{}^\circ C\]
Correct Answer: A
Solution :
Mass of solution = 100 g, Mass of glucose =1g, Mass of solvent \[=100-1=99\text{ }g\] w = mass of solute glucose, W = mass of solvent m = molar mass of solute \[\Delta {{T}_{f}}={{K}_{f}}\times \frac{w}{m}\times \frac{1}{W/1000}\] \[=1.84\times \frac{1}{180}\times \frac{1}{99/1000}=0.103\] Freezing point of the solution \[=273-0.103\] \[=272.897\text{ }K\]
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