A) 1.12 K
B) -0.56
C) 0.56 K
D) -1.12K
Correct Answer: A
Solution :
\[\Delta \Tau =i{{K}_{f}}m\] We know, \[\alpha =\frac{i-1}{n-1}\](dissociation) \[\alpha =\frac{i-1}{2-1}\] \[i=\alpha +1\] so \[\Delta {{\Tau }_{f}}=(\alpha +1){{K}_{f}}m\] Given \[\alpha =0.2,\]molality \[=0.5,K_{f}^{'}=1.86\] \[\therefore \]\[\Delta T=0.5\times 1.2\times 1.86=1.116\,K\]You need to login to perform this action.
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