Boiling point of water at 750 mm Hg is 96.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? Molal elevation constant for water is 0.52 K kg mol-1.

Question

Boiling point of water at 750 mm Hg is 96.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? Molal elevation constant for water is 0.52 K kg mol-1. (A) 11.08 kg (B) 1108.2 g (C) 1108.2 kg (D) 11.08 g

Tardigrade Admin · Accepted Answer

Elevation in boiling point (ÎTb) = Tb - Tb° = 100 - 96.63 = 3.37°C Mass of solvent (water), W1 = 500 g Molar mass of solvent, M1 = 18 g mol-1 Molar mass of solute, (C12H22O11), M2 = 342 g mol-1 Mass of sucrose to be added (W2) = ? M2 = (1000 Kb W2/W1 Δ Tb) From formula, W2 = (M2× W1× Δ Tb/1000× Kb ) = (342× 500× 3.37/1000× 0.52) = 1108.2 g

Q. Boiling point of water at 750 mm Hg is 96.63∘C. How much sucrose is to be added to 500 g of water such that it boils at 100∘C? Molal elevation constant for water is 0.52Kkgmol−1.

11.08 kg

1108.2 g

1108.2 kg

11.08 g

Q. Boiling point of water at 750 mm Hg is $96.6 3^{\circ} C$ . How much sucrose is to be added to 500 g of water such that it boils at $10 0^{\circ} C$ ? Molal elevation constant for water is $0.52 K k g m o l^{- 1}$ .