0.01 M solution of KCl and BaCl2 are prepared in water. The freezing points of KCl is found to be -2° C . What is the freezing point of BaCl2 solution assuming both KCl and BaCl2 to be completely ionized?

Question

0.01 M solution of KCl and BaCl2 are prepared in water. The freezing points of KCl is found to be -2° C . What is the freezing point of BaCl2 solution assuming both KCl and BaCl2 to be completely ionized? (A)  -3° C  (B)  +3° C  (C)  -2° C  (D)  -4° C

Tardigrade Admin · Accepted Answer

i for KCl=2,i for BaCl2=3 ∵ Δ Tf propto i (Δ Tf(KCl)/Δ Tf(BaCl2))=(2/3) Δ Tf(BaCl2)=(3/2)× 2=3° C ∴ freezing point of KCl=-3° C

Q. 0.01 M solution of $K Cl$ and $B a C l_{2}$ are prepared in water. The freezing points of $K Cl$ is found to be $- 2^{\circ} C$ . What is the freezing point of $B a C l_{2}$ solution assuming both $K Cl$ and $B a C l_{2}$ to be completely ionized?

$- 3^{\circ} C$

$+ 3^{\circ} C$

$- 2^{\circ} C$

$- 4^{\circ} C$

$5^{\circ} C$

$i$ for $K Cl = 2, i$ for $B a C l_{2} = 3$ $∵$ $Δ T_{f} \propto i$ $\frac{Δ T _{f} ( K Cl )}{Δ T _{f} ( B a C l _{2} )} = \frac{2}{3}$ $Δ T_{f} (B a C l_{2}) = \frac{3}{2} \times 2 = 3^{\circ} C$ $∴$ freezing point of $K Cl = - 3^{\circ} C$

Q. 0.01 M solution of KCl and BaCl2​ are prepared in water. The freezing points of KCl is found to be −2∘C . What is the freezing point of BaCl2​ solution assuming both KCl and BaCl2​ to be completely ionized?

−3∘C

+3∘C

−2∘C

−4∘C

5∘C

i for KCl=2,i for BaCl2​=3 ∵ ΔTf​∝i ΔTf​(BaCl2​)ΔTf​(KCl)​=32​ ΔTf​(BaCl2​)=23​×2=3∘C ∴ freezing point of KCl=−3∘C