Question

Boiling point of water at 750 mm Hg is 99.63°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.

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C?

Answer 1

Here, elevation of boiling point (ΔT_b) = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water (w₁) = 500 g

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂ = 11 × 12 + 22 × 1 + 11 × 16

 = 342 g mol⁻¹

Molal elevation constant (K_b) = 0.52 K kg mol⁻¹

We know that,

ΔT_b = `(K_b xx 1000 xx w_2)/(M_2 xx w_1)`

⇒ w₂ = `(Delta T_b xx M_2 xx w_1)/(K_b xx 1000)`

= `(0.37 xx 342 xx 500)/(0.52 xx 1000)`

= 121.67 g (approximately)

= 122 g

Hence, 122 g of sucrose is to be added.