Solutions Question 347
Question: A solution containing 0.85 g of $ ZnCl_2 $ in 125.0 g of water freezes at $ -0.23{}^\circ C $ . The apparent degree of dissociation of the salt is ( $ K_{f} $ for water $ =1.86Kkgmo{l^{-1}} $ , atomic mass: Zn = 65.3 and Cl = 35.5)
Options:
A) 1.36%
B) 73.5%
C) 7.35%
D) 2.47%
Show Answer
Answer:
Correct Answer: B
Solution:
$ 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-( -23{}^\circ C )=23{}^\circ C $ Now, $ i=\frac{M_{normal}}{M_{observed}}=\frac{136.3}{55}=2.47 $
$ \underset{1-\alpha }{\mathop{ZnCl_2}},\rightarrow \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% $
