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Question

# To 500cm3 of water 3.0×10−3kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression of freezing point? [Kf and density of water are 1.86Kkgmol−1 and 0.997gcm−3, respectively]

A
0.23K
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B
0.33 K
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C
0.98 K
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D
0.68 K
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Solution

## The correct option is C 0.23KMolar mass of acetic acid is 60g/mol. Number of moles of acetic acid =massmolarmass=360=0.05 moles. Density of water is 0.997 g/ml. 500 ml of water corresponds to 500×0.997=498.5g or 0.4985 kgMolality is the number of moles of acetic acid divided by the mass of water (in kg).Molality m=0.050.4985=0.1m23 % dissociation of acetic acid means degree of dissociation α=0.23The vant Hoff factor i=[1+(n−1)α]=[1+(2−1)×0.23]=1.23The depression in the freezing point ΔTf=i×kf×m=1.23×1.86×0.1=0.23KHence, the correct option is A

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