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JEE Advanced levelSingle correct optionDifficulty 3 of 3so104

A weak monobasic acid HA (molar mass $60 g mo l^{- 1}$ ), $3.0 g$ , is dissolved in $250 g$ of water and the boiling point rises by $0.1248 K$ ( $K_{b} = 0.52 K kg mo l^{- 1}$ ). Using $i = 1 + α$ for $HA H X^{+} + A X^{-}$ , the degree of dissociation of HA is

A $0.40$
B $0.20$
C $0.10$
D $1.20$

Answer: B

Calculated

m = (3.0/60) /0.250 = 0.20 mol k g^{- 1}

, so

Δ T_{b} (calc) = 0.52 \times 0.20 = 0.104 K

.

i = 0.1248/0.104 = 1.20

. With

i = 1 + α

,

α = 0.20

.

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