Answer :

Given: Mass of CH3CH2CHClCOOH = 10 g


Mass of water = 250g


Ka = 1.4 × 10–3,


Kf = 1.86 K kg mol–1


Molar mass of CH3CH2CHClCOOH = 12 + 3 + 12 + 2 + 12 + 1 + 35.5 + 2 + 16 + 16 + 1


= 122.5 g mol–1


Number of moles of solute =


No. of moles =


No. of moles = mol


Now, Molality is given as,




M = 0.3264 kg/mol


CH3CH2CHClCOOH CH3CH2CHClCOO- + H +



Total moles at equilibrium = (1-α) + 2 α


= 1 + α


In order to find out the depression in freezing point, ff values of i(vant Hoff’s factor) and α(degree of dissociation) are to be found out.


To find out degree of dissociation, α



Here, the value of α is negligible as compared to 1, and hence 1- = 1, giving,






To find out Vant Hoff’s factor,






Now, to find out the depression in freezing point,


ff





Thus, the depression in freezing point is f = 0.650C


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