Solutions Question 207
Question: Freezing point of an aqueous solution is $ -0.186{}^\circ C. $ If the values of $ K_{b} $ and $ K_{f} $ of water are respectively $ 0.52Kkgmo{l^{-1}} $ and $ 1.86Kkgmo{l^{-1}} $ , then the elevation of boiling point of the solution in K is
Options:
A) 0.52
B) 1.04
C) 1.34
D) 0.052
Show Answer
Answer:
Correct Answer: D
Solution:
$ \Delta T_{f}=i.K_{f}.m;\Delta T_{b}=i.K_{b}.m $
$ \frac{\Delta T_{f}}{\Delta T_{b}}=\frac{K_{f}}{K_{b}} $
$ \Delta T_{f}=0-( -0.186{}^\circ C )=0.186{}^\circ C $
$ \frac{0.186}{\Delta T_{b}}=\frac{1.86}{0.52}\Rightarrow \Delta T_{b}=\frac{0.52\times 0.186}{1.86}=0.052 $
BETA
