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NCERT Class XII Chemistry
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Question : 32 of 53
Marks: +1, -0
Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH\mathrm{CH_3CH_2CHClCOOH} is added to 250 g of water. (Ka=1.4×10−3,Kf=1.86 K kg mol−1)(K_a = 1.4 \times 10^{-3}, K_f = 1.86 \, \mathrm{K \, kg \, mol^{-1}})
Solution:  
Number of moles of CH3CH2CHClCOOH\mathrm{CH_3CH_2CHClCOOH} =10122.5= \frac{10}{122.5} mole =8.16×10−2=8.16 \times 10^{-2} mole
∴ Molality of the solution (m) =8.16×10−2 mol250×1000 kg−1= \frac{8.16 \times 10^{-2} \, \mathrm{mol}}{250} \times 1000 \, \mathrm{kg}^{-1} =0.3264 m=0.3264 \, \mathrm{m}
If α is the degree of dissociation of CH3CH2CHClCOOH\mathrm{CH_3CH_2CHClCOOH}, then
CH3CH2CHClCOOH⇌CH3CH2CHClCOO−+H+Initial conc.C mol L−100At equilibriumC(1−α)CαCα\begin{array}{lccccc} & \mathrm{CH_3CH_2CHClCOOH} & \rightleftharpoons & \mathrm{CH_3CH_2CHClCOO}^- & + & \mathrm{H}^+ \\ \text{Initial conc.} & C \, \mathrm{mol \, L}^{-1} & & 0 & & 0 \\ \text{At equilibrium} & C(1 - \alpha) & & C\alpha & & C\alpha \end{array}
∴Ka=Cα×CαC(1−α)=Cα2\therefore K_a = \frac{C\alpha \times C\alpha}{C(1-\alpha)} = C\alpha^2 or, α=KaC\alpha = \sqrt{ \frac{K_a}{C} } =1.4×10−30.3264=0.065= \sqrt{ \frac{1.4 \times 10^{-3}}{0.3264} } = 0.065
To calculate van’t Hoff factor:
CH3CH2CHClCOOH⇌CH3CH2CHClCOO−+H+Initial conc.100At equilibrium1−ααα\begin{array}{lccccc} & \mathrm{CH_3CH_2CHClCOOH} & \rightleftharpoons & \mathrm{CH_3CH_2CHClCOO}^- & + & \mathrm{H}^+ \\ \text{Initial conc.} & 1 & & 0 & & 0 \\ \text{At equilibrium} & 1 - \alpha & & \alpha & & \alpha \end{array}
Total moles = 1 + a
∴ i = 1 + 0.065 = 1.065
ΔTf=iKfm\Delta T_f = i K_f m = (1.065)(1.86)(0.3264) = 0.65°
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