Test Index

CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

© examsnet.com
Question : 28 of 37
Marks: +1, -0
SECTION -C
A 0.01m0.01 \text{m} aqueous solution of AlCl3\mathrm{AlCl}_3 freezes at 0.068∘C0.068^{\circ} \text{C}. Calculate the percentage of dissociation.
[Given : KfK_f for Water =1.68Kkgmol−1=1.68 \text{K} \text{kg} \text{mol}^{-1} ]
Solution:  
Given, m=0.01mm=0.01 \text{m}
  ΔTf(s)=−0.068∘C\; \Delta T_f(\text{s}) = -0.068^{\circ} \text{C}
  Kf(aq)=1.86  K  kg  mol−1\; K_f(\text{aq}) = 1.86 \; \text{K} \; \text{kg} \; \text{mol}^{-1}
  ΔTf=iKfm\; \Delta T_f = i K_f m
  i=ΔTfKf×m\; i = \frac{\Delta T_f}{K_f} \times m
  i=0.0681.86×0.01 m=3.65\; i = \frac{0.068}{1.86} \times 0.01 \text{ m} = 3.65
AlCl3→Al3++3Cl−initial1 mol00At equilibrium1−αα3α\begin{array}{c c c c c c} {} & \mathrm{AlCl}_3 & \rightarrow & \mathrm{Al}^{3+} & + & 3\mathrm{Cl}^{-} \\ \text{initial} & 1\text{ mol} & {} & 0 & {} & 0 \\ \text{At equilibrium} & 1-\alpha & {} & \alpha & {} & 3\alpha \end{array}
Total number of moles at equilibrium
=1−α+α+3α=1+3α= 1-\alpha+\alpha+3\alpha = 1+3\alpha
l=Total no. of moles at equilibriumInitial no. of molesl = \frac{ \text{Total no. of moles at equilibrium} }{ \text{Initial no. of moles} }
=1+3α1= \frac{1+3\alpha}{1}
3.65=1+3α3.65 = 1 + 3\alpha
α=3.65−13\alpha = \frac{3.65 - 1}{3}
Percentage dissociation =0.88%= 0.88\%.
© examsnet.com
Go to Question:
Prev Question
Next Question