A) \[-3{}^\circ C\]
B) \[+3{}^\circ C\]
C) \[-2{}^\circ C\]
D) \[-4{}^\circ C\]
E) \[5{}^\circ C\]
Correct Answer: A
Solution :
\[i\]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 \[KCl=-3{}^\circ C\]You need to login to perform this action.
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