CBSE Class 12 · 2004 · All India Compartment · SolutionsShort Answer

45 g of ethylene glycol (C

_{2}

_{6}

_{2}

) is mixed with 600 g of water. Calculate:
(a) the freezing point depression,
(b) the freezing point of the solution. (K

_{f}

for water = 1.86 K kg mol

^{- 1}

; atomic masses: C = 12, H = 1, O = 16 u)

Answer: ΔT_f ≈ 2.25 K; T_f ≈ −2.25°C (270.9 K)

Solution

Molar mass of ethylene glycol = 62 g mol

^{- 1}

.
Molality:

m = \frac{45/62}{0.600} = \frac{0.7258}{0.600} = 1.210 mol kg^{- 1}

(a) Depression of freezing point:

Δ T_{f} = K_{f} m = 1.86 \times 1.210

Δ T_{f} = 2.25 K

(b) Freezing point:

T_{f} = 0 - 2.25 = - 2.25^{\circ} C (= 270.90 K)

CBSE Class 12 Board · 2004 · All India Compartment · Subjective