A) \[-{{3}^{o}}C\]
B) \[+{{3}^{o}}C\]
C) \[-{{2}^{o}}C\]
D) \[-{{4}^{o}}C\]
Correct Answer: A
Solution :
\[i\]for\[KCl=2,\]for for\[CaC{{L}_{2}}=3\], \[\Delta {{T}_{f}}\propto i\] \[\frac{\Delta {{T}_{f}}(KCl)}{\Delta {{T}_{f}}(CaC{{l}_{2}})}=\frac{2}{3}\] \[\Delta {{T}_{f}}(CaC{{l}_{2}})=\frac{3}{2}\times 2={{3}^{o}}C\] Freezing point of\[CaC{{l}_{2}}={{3}^{o}}C\]
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