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Chemistry Question 55 – JEE-MAIN 2026

19.5 g of fluoro acetic acid (molar mass = 78 g mol1) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1°C. What is Ka of fluoro acetic acid ? (For water, Kf=1.86 K kg mol1). Assume molarity and molality to have same values.

The depression in freezing point depends on the total number of particles in the solution, which is affected by the dissociation of the weak acid.

Step 1: Calculate Molality and Van't Hoff Factor✦ Active

First, calculate the moles of fluoro acetic acid and the molality of the solution. Then, use the freezing point depression formula to find the Van't Hoff factor (i). Moles of fluoro acetic acid (n2) = MassMolar Mass=19.5 g78 g/mol=0.25 mol. Mass of water (w1) = 500 g=0.5 kg. Molality (m) = n2w1(kg)=0.25 mol0.5 kg=0.5 mol/kg. Using the freezing point depression formula: ΔTf=iKfm.

1=i×1.86 K kg mol1×0.5 mol/kg 1=i×0.93 i=10.931.07526
Step 2: Determine the Degree of Dissociation (α)○ Expand

For a weak acid (HFA) that dissociates into two ions (H+ and FA), the Van't Hoff factor (i) is related to the degree of dissociation (α) by the formula i=1+α.

α=i1 α=1.075261=0.07526
Step 3: Calculate the Acid Dissociation Constant (Ka)○ Expand

The acid dissociation constant (Ka) for a weak acid is given by the expression Ka=Cα21α, where C is the initial concentration (molality in this case, as molarity and molality are assumed to be same).

Ka=0.5×(0.07526)210.07526 Ka=0.5×0.0056640.92474 Ka=0.0028320.924740.003062

Rounding this value, Ka3×103.

💡 Teacher's Secret Hint

Ensure to use the correct formula for Ka in terms of C and α for a weak acid.

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