Question

0.01 M solution of $ KCl $ and $ BaC{{l}_{2}} $ are prepared in water. The freezing points of $ KCl $ is found to be $ -2{}^\circ C $ . What is the freezing point of $ BaC{{l}_{2}} $ solution assuming both $ KCl $ and $ BaC{{l}_{2}} $ to be completely ionized?

• A. $ -3{}^\circ C $
• B. $ +3{}^\circ C $
• C. $ -2{}^\circ C $
• D. $ -4{}^\circ C $
• E. $ 5{}^\circ C $

Answer 1

Answer: (A)

$ i $ for $ KCl=2,i $ for $ BaC{{l}_{2}}=3 $ $ \because $ $ \Delta {{T}_{f}}\propto i $ $ \frac{\Delta {{T}_{f}}(KCl)}{\Delta {{T}_{f}}(BaC{{l}_{2}})}=\frac{2}{3} $ $ \Delta {{T}_{f}}(BaC{{l}_{2}})=\frac{3}{2}\times 2=3{}^\circ C $ $ \therefore $ freezing point of $ KCl=-3{}^\circ C $