A 0.20 m aqueous solution of K2SO4 is 80% dissociated (i = 1 + (n-1)alpha). With K_f = 1.86 K kg mol^-1, the depression of freezing point is

JEE Main levelSingle correct optionDifficulty 2 of 3so097

A $0.20 m$ aqueous solution of $K X_{2} SO X_{4}$ is $80%$ dissociated ( $i = 1 + (n - 1) α$ ). With $K_{f} = 1.86 K kg mo l^{- 1}$ , the depression of freezing point is

n = 3

, so

i = 1 + 2 \times 0.80 = 2.60

Δ T_{f} = i K_{f} m = 2.60 \times 1.86 \times 0.20 = 0.967 K