Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1.
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,
Thus, the depression in freezing point is f = 0.650C
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