A) \[-{{0.480}^{o}}C\]
B) \[-{{0.360}^{o}}C\]
C) \[-{{0.260}^{o}}C\]
D) \[+{{0.480}^{o}}C\]
Correct Answer: A
Solution :
\[\Delta {{T}_{f}}=molarity\times {{k}_{f}}\times i\] \[HX{{H}^{+}}+{{X}^{-1}}\] \[=0.2\times 1.85\times 1.3\] \[\alpha =0.3\] \[={{0.481}^{o}}\] \[i=1+\alpha =1.3\] \[\therefore \]\[f.p.={{0.481}^{o}}C\]You need to login to perform this action.
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