  Question

Question 2.32 Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is added to 250 g of water, Ka=1.4×10−3; Kf=1.86 K kg mol−1

Solution

Step I: Calculation of degree of dissociation Mass of solute = 10g Molar mass of solute (CH3CH2CHClCOOH)=(12×4)+(1×7)+(35.5)+(16×2)=48+7+35.5+32=122.5 g mol−1 Molal concentration of solution =MassMolar massMassofsolvent (kg)=(10g)(122.5 g mol−1)×(0.25 kg)=0.326m If α is the degree of dissociation of CH3CH2CHClCOOH, then CH3CH2CHClCOOH⇌CH3CHClCOO−+H+Initial conc. C mol−1 kg00At equilibrium C(1−α)CαCα ∴Ka=Cα×CαC(1−α)=Cα2 [Considering (1−α)=1 or dilute solutions] α2=KaC or α=√KaC So, α=√1.4×10−30.326=√42.9×10−4=6.55×10−2=0.065 Step II: Calculation of van't Hoff factor (f) CH3CH2CHClCOOH⇌CH3CH2CHClCOO−+H+Initial no. of moles100Moles at equilibrium1−ααα Total number of moles after dissociation =(1−α)+(α)+(α)=(1+α) Van't Hoff fraction (i) =Total number of moles after dissociationNumber of moles before dissociation i=(1+α)1=(1+α) =1+0.065=1.065 Step III: Calculation of depression in freezing point (ΔTf) ΔTf=iKfm=(1.065)×(1.86K kg mol−1)×(0.326 mol kg−1) ΔTf=0.65K ChemistryNCERT TextbookStandard XII

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