Solutions Question 202
Question: Freezing point of an aqueous solution is $ ( -0.186 ){}^\circ C $ . Elevation of boiling point of the same solution is $ K_{b}=0.512{}^\circ C,K{ _{f}}=1.86{}^\circ C $ , find the increase in boiling point.
Options:
A) $ 0.186{}^\circ C $
B) $ 0.0512{}^\circ C $
C) $ 0.092{}^\circ C $
D) $ 0.2372{}^\circ C $
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Answer:
Correct Answer: B
Solution:
$ \Delta T_{b}=K_{b}\frac{W_{B}}{M_{B}\times W_{A}}\times 1000; $
$ \Delta T_{f}=K_{f}\frac{W_{B}}{M_{B}\times W_{A}}\times 1000; $
$ \frac{\Delta T_{b}}{\Delta T_{f}}=\frac{K_{b}}{K_{f}}=\frac{\Delta T_{b}}{-0.186}=\frac{0.512}{1.86}=0.0512{}^\circ C. $
BETA
