0.01 M solution of KCl and CaCl 2 are prepared in water. The freezing point of KCl is found to be -2° C . What is the freezing point of CaCl 2 to be completely ionised?

Question

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

Tardigrade Admin · Accepted Answer

i for K C l=2i, for for C a C l2=3 Δ Tf propto i (Δ Tf(K C l)/Δ Tf(C a C l2))=(2/3) Δ Tf(CaCl2)=(3/2) × 2=3° C Freezing point of CaCl2=3° C

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

$- 3^{\circ} C$

$+ 3^{\circ} C$

$- 2^{\circ} C$

$- 4^{\circ} C$

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

Q. 0.01M solution of KCl and CaCl2​ are prepared in water. The freezing point of KCl is found to be −2∘C. What is the freezing point of CaCl2​ to be completely ionised?

−3∘C

+3∘C

−2∘C

−4∘C