A) 272.898 K
B) \[~0.102{{\,}^{o}}C\]
C) 273 K
D) \[108{{\,}^{o}}C\]
Correct Answer: A
Solution :
Mass of solution \[=100\text{ }g,\] Mass of glucose \[=1\text{ }g,\] Mass of solvent\[~=100-1=99\text{ }g\] \[\text{w}\,\text{=}\]mass of solute glucose,\[\text{W =}\]mass of solvent \[\text{m =}\]molar mass of solute \[\Delta {{T}_{f}}\times \frac{w}{m}\times \frac{1}{W/1000}\] \[=1.84\times \frac{1}{180}\times \frac{1}{99/1000}=0.103\] Freezing point of the solution \[=273-0.103\] \[=272.897\text{ }K\]
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