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← JEE Main 2026 · 2 Apr · Shift 1
ChemistrySingle correct+4 / -1Question 55 of 75

JEE Main 2026 · 2 Apr · Shift 1, Chemistry question 55

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

Choose one option
Answer and worked solution

Official answer: D

Molality. 19.5/780.5=0.5\frac{19.5/78}{0.5}=0.5 mol kg⁻¹.

Van 't Hoff factor. i=ΔTfKfm=11.86×0.5=1.075i=\dfrac{\Delta T_f}{K_fm}=\dfrac1{1.86\times0.5}=1.075, so the degree of dissociation is α=0.075\alpha=0.075.

Ka. Ka=cα21−α=0.5(0.075)20.925≈3×10−3K_a=\dfrac{c\alpha^2}{1-\alpha}=\dfrac{0.5(0.075)^2}{0.925}\approx3\times10^{-3}.

(D)