A) \[-{{0.31}^{o}}C\]
B) \[-{{0.45}^{o}}C\]
C) \[-{{0.53}^{o}}C\]
D) \[-{{0.90}^{o}}C\]
Correct Answer: B
Solution :
\[\Delta {{T}_{f}}=\text{molality}\times {{K}_{f}}\times (1+\alpha )\] \[\alpha =0.2\], Molality = 0.2, \[{{K}_{f}}=1.86\] \[\Delta {{T}_{f}}=0.2\times 1.2\times 1.86={{0.4464}^{o}}\] Freezing point = \[-{{0.45}^{o}}C\].
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