A) 40
B) 50
C) 60
D) 80
Correct Answer: A
Solution :
Given, molality of solution, \[m=0.5\,m\] \[\Delta {{T}_{f}}=2K\] \[{{K}_{f}}=5K\,kg\,mo{{l}^{-1}}\] \[%\]of association, \[\alpha =?\] Step I Calculation of van?t Hoff factor (i) \[\Delta {{T}_{f}}=i{{K}_{f}}M\] \[i=\frac{\Delta {{T}_{f}}}{{{K}_{f}}M}\] \[i=\frac{2K}{5\,K\,kg\,mo{{l}^{-1}}\times 0.5\,m}\] \[i=\frac{4}{5}\] Step II Calculation of % association of acid \[(\alpha ).\] As solute form dimer in benzene, i.e. \[2A{{A}_{2}}\] Therefore, value of \[x=2\] \[\alpha =\frac{i-1}{1/x-1}\] \[\alpha =\frac{4/5-1}{1/2-1}=0.4\] \[\therefore \] % of \[\alpha =0.4\times 100=40%\]You need to login to perform this action.
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