A) \[-3{}^\circ C\]
B) \[-2{}^\circ C\]
C) \[3{}^\circ C\]
D) \[2{}^\circ C\]
E) None of these
Correct Answer: A
Solution :
\[i\]for \[KCl=2\]\[i\]for\[BaC{{l}_{2}}=3\] \[\frac{\Delta {{T}_{f}}(KCl)}{\Delta {{T}_{f}}(BaC{{l}_{2}})}=\frac{2}{3}\] \[\therefore \] \[\Delta {{T}_{f}}(BaC{{l}_{2}})=\frac{3}{2}\times \Delta {{T}_{f}}(KCl)\] \[=\frac{3}{2}\times 2=3{}^\circ \] \[\therefore \] Freezing point of\[KCl=-3{}^\circ C\]You need to login to perform this action.
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