A) 40
B) 50
C) 60
D) 80
Correct Answer: A
Solution :
Given, molality of solution,\[m=0.6\text{ }m\] \[\Delta {{T}_{f}}=2K\] \[{{K}_{f}}=5\,K\,kg\,mo{{l}^{-1}}\] % of association, \[\alpha =?\] Step I Calculation of vant 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 . 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 \[a=0.4\times 100=40%\]
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