2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water =0.52 K kg mol⁻¹ )

Solution

Chemistry ma’am · 3 steps

Both solutes are non-electrolytes, so add their moles together and treat them as one solute.

1. Count solute particles. Ethylene glycol and glucose do not ionise in water, so i = 1 for each. Colligative properties depend only on the total number of solute particles, so the two amounts add up: 2 + 2 = 4 mol.
2. Find molality. The water is 500 g = 0.5 kg, so m = 4/0.5 = 8 mol kg⁻¹.

   Δ T_b = K_b × m = 0.52 × 8 = 4.16 K

3. So. The solution boils at 373.15 + 4.16 = 377.31 K ≈ 377.3 K, which is option (a).

The idea to keep. In a mixture of non-electrolytes, every solute adds to the colligative effect. Add all their moles before you find molality, then add Δ T_b to the boiling point of water, 373 K.

Answer from the NTA final answer key. Solution by PrepWiser.