A) 271.326 K
B) 272 K
C) 270.5 K
D) 268.5 K
E) 269 K
Correct Answer: A
Solution :
\[\Delta {{T}_{f}}=i\times {{k}_{f}}\times \]molality \[KBr\]is dissociated as: \[KBr\rightleftharpoons {{K}^{+}}+B{{r}^{-}}\] Here, degree of dissociation = 80% \[\because \] \[i=\frac{1-\alpha +n\alpha }{1}\]where n = number of ions produced by the complete dissociation \[i=\frac{1-0.8+2\times 0.8}{1}=1.8\] \[\therefore \] \[\Delta {{T}_{f}}=1.8\times 1.86\times 0.5\] \[\Delta {{T}_{f}}=1.674\,K\] \[\Delta {{T}_{f}}=T{{{}^\circ }_{f}}-{{T}_{f}}\] \[1.674=273-{{T}_{f}}\] \[{{T}_{f}}=273-1.674\] \[{{T}_{f}}=271.326\,K\]
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