A) \[-{{2}^{o}}C\]
B) \[-{{3}^{o}}C\]
C) \[-{{1.5}^{o}}C\]
D) \[-{{1.66}^{o}}C\]
Correct Answer: B
Solution :
\[i=NaCl=2,\,i\] for \[BaC{{l}_{2}}=3\] Hence, \[\frac{\Delta {{T}_{f}}(NaCl)}{\Delta {{T}_{f}}(BaC{{l}_{2}})}=\frac{2}{3}\] or \[{{T}_{f}}(BaC{{l}_{2}})=\frac{3}{2}\times 2={{3}^{o}}\] \[(\Delta {{T}_{f}}NaCl)=2)\] So that \[{{T}_{f}}\] for Bad \[BaC{{l}_{2}}=-{{3}^{o}}C\]
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