A) 1.36%
B) 73.5%
C) 7.35%
D) 2.47%
Correct Answer: B
Solution :
[b] \[Mol.wt.=\frac{{{K}_{f}}\times w\times 1000}{\Delta {{T}_{f}}\times W}\] \[=\frac{1.86\times 0.85\times 1000}{0.23\times 125}\approx 55g\] Where \[w=0.85g\] \[W=125g\] \[\Delta {{T}_{f}}=0{}^\circ C-\left( -23{}^\circ C \right)=23{}^\circ C\] Now, \[i=\frac{{{M}_{normal}}}{{{M}_{observed}}}=\frac{136.3}{55}=2.47\] \[\underset{1-\alpha }{\mathop{ZnC{{l}_{2}}}}\,\rightleftharpoons \underset{\alpha }{\mathop{Zn}}\,{{}^{++}}+2\underset{2\alpha }{\mathop{C{{l}^{-}}}}\,\] Van't Hoff factor (i) \[=\frac{1-\alpha +\alpha +2\alpha }{1}=2.47\] \[\therefore \,\,\,\,\,\alpha =0.735=73.5%\]
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