To 500 cm3 of water 3.0 × 10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? Kf and density of water are 1.86 K kg mol–1 and 0.997g cm–3,
respectively.
Text Solution
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Number of moles of acetic acid= 60gmol−13×10−3kg=60gmol−13g=0.05
Mass of water= m1=500cm3×0.997gcm−3=0.4985kg
Molality of acetic acid= m=m1n2=0.4985kg0.05mol=0.1003molkg−1
Dissociation of CH3COOH :
CH3COOH ⇌ CH3COO ⊝ +H ⊕
t=0 1 0 0
eqm 1−α α α
Total number of moles= i=1−α+α+α=1+α
as, ΔTf=iKfm
ΔT=(1+α)Kfm=(1+0.23)(1.86 K Kg mol−1)(0.1003 mol Kg−1)=0.228K
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