Question

18 g of Glucose (C₂H₁₂O₆) was dissolved in 1 kg of water at 1.013 bar atmospheric pressure. At which temperature will water boil?

K_b of water is 0.52 Kg mol^-1

Answer 1

From question W = 18g, M₀ = 1 Kg = 1000g, T₁ = 100°C, T₂ = ?

x = Molecular mass of glucose (C₂H₁₂O₆)

= 6 x 12 + 12 x 1 + 6 x 16 = 72 + 12 + 96 = 180

we know that,

Again ΔT_b = T₂ - T₁

∴ T₂ = ΔT_b + T₁ = 0.052 + 100 = 100.052°C

Thus, water will boil at = 100.052°C