If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water $(\Delta \mathrm{T}_{\mathrm{f}})$, when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is ( $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

JEE Main · 2010 · 11 Apr 2010 · Offline · Solutions

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water

(Δ T_{f})

, when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (

K_{f} = 1.86 K kg mol^{- 1}

)

(A)0.0372 K
(B)0.0558 K✓ Correct
(C)0.0744 K
(D)0.0186 K

Answer: (B)

Solution

Taken as fully dissociated, sodium sulphate gives three ions per formula unit,

2 N a^{+}

and one

S O_{4}^{2 -}

, so the van't Hoff factor is

i = 3

. The freezing-point depression is

Δ T_{f} = i K_{f} m

, and with

m = 0.01

mol kg

^{- 1}

and

K_{f} = 1.86

K kg mol

^{- 1}

Δ T_{f} = 3 \times 1.86 \times 0.01 = 0.0558

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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