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CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

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Question : 29 of 35
Marks: +1, -0
When 19.5 g19.5\ \mathrm{g} of F−CH2−COOH\mathrm{F}-\mathrm{CH}_2-\mathrm{COOH} (Molar mass =78 g mol−1=78\ \mathrm{g}\ \mathrm{mol}^{-1} ), is dissolved in 500 g500\ \mathrm{g} of water, the depression in freezing point is observed to be 1∘C1^{\circ}\mathrm{C}.
Calculate the degree of dissociation of F−CH2−\mathrm{F}-\mathrm{CH}_2- COOH.
[Given : KfK_f for water =1.86 K kg mol−1=1.86\ \mathrm{K}\ \mathrm{kg}\ \mathrm{mol}^{-1} ]
Solution:     👈: Video Solution
Molecular mass of 78 g/mol78\ \mathrm{g}/\mathrm{mol}
(F−CH2 COOH)(\mathrm{F}-\mathrm{CH}_2\ \mathrm{COOH})
No. of moles of fluoroacetic acid is 19.578=0.25\frac{19.5}{78}=0.25
Molality is the number of moles of solute in 1 kg1\ \mathrm{kg} of solvent
Molality =0.255001000=0.50 m=\frac{0.25}{\frac{500}{1000}}=0.50\ \mathrm{m}
ΔTf=Kf×m=1.86×0.50\Delta T_f = K_f \times m = 1.86 \times 0.50 =0.93 K=0.93\ \mathrm{K}
i=1.00.93=1.0753i = \frac{1.0}{0.93} = 1.0753
CH2FCOOH→CH3FCOO−+H+\mathrm{CH}_2\mathrm{FCOOH} \rightarrow \mathrm{CH}_3\mathrm{FCOO}^{-} + \mathrm{H}^{+}
C(1−α)  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;Cα  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;  ;CαC(1-\alpha)\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;C\alpha\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;\;;C\alpha
Total number of moles =C(1−α)+Cα+Cα=C(1-\alpha)+C\alpha+C\alpha =C(1+α)=C(1+\alpha)
i=C(1+α)C=1+α=1.0753i = \frac{C(1+\alpha)}{C} = 1+\alpha = 1.0753
α=0.0753\alpha = 0.0753
(ii) (α=0.50×0.0753=0.03765(\alpha = 0.50 \times 0.0753 = 0.03765
C(1−α)=0.50(1−0.0753)=0.462C(1-\alpha)=0.50(1-0.0753)=0.462
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