NEETprep Answer:
Molar mass of CH3CH2CHClCOOH=15+14+13+35.5+12+16+16+1=122.5 g mol-1
∴No. of moles present in 10 g of CH3CH2CHClCOOH=10 g122.5 g mol-1CH3CH2CHCICOOH=10 g122.5 g mol-1
=0.0816 mol
It is given that 10 g of is added to 250 g of water.
Molality of the solution,
Let α be the degree of dissociation of CH3CH2CHClCOOH.
CH3CH2CHClCOOH undergoes dissociation according to the following equation:
CH3CH2CHCICOOH↔CH3CH2CHCICOOH-+H+
Initial conc. C mol L-1 0 0
At equlibrium C(1-a) Ca Ca
∴ Ka=Ca.CaC(1-a)
=Ca21-a
Since α is very small with respect to 1, 1 − α ≈ 1
Now, Ka=Ca21
⇒Ka=Cα2
⇒a=KaC
=1.4×10-30.3264 (∴Ka=1.4×10-3)
=0.0655
Again,
CH3CH2CHClCOOH↔CH3CH2CHClCOOH-+H+
Initial conc. 1 0 0
At equilibrium 1-α α α
Total moles of equilibrium = 1 − α + α + α = 1 + α
∴i=1+α1
=1+α
=1+0.0655
=1.0655
Hence, the depression in the freezing point of water is given as:
∆Tf=i.Kfm
=1.0665×1.86 K kg mol-1×0.3264 mol kg-1
=0.65 K
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