Question

A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K

Given : (Molar mass of sucrose = 342 g mol⁻¹)

(Molar mass of glucose = 180 g mol⁻¹)

Answer 1

Here, ΔT_f = (273.15 − 269.15) K = 4 K

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol⁻¹

10% solution (by mass) of sucrose (cane sugar) in water means 10 g of cane sugar is present in (100 − 10)g = 90 g of water.

Now, number of moles of cane sugar = `10/342 = 0.0292 mol`

Therefore, molality (m) of the solution, = `(0.0292 xx 1000)/90 = 0.3244 mol kg^(-1)`

Applying the relation,

ΔT_f = K_f × m

`=> K_f = (triangleT_t)/m = 4/0.3244 = 12.33 "K Kg mol"^(-1)`

Molar mass of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol⁻¹

10% glucose in water means 10 g of glucose is present in (100 − 10) g = 90 g of water.

∴ Number of moles of glucose = `10/180` mol

= 0.0555 mol

Therefore, molality (m) of the solution, = `(0.0555 xx 1000)/90` = 0.6166 mol kg⁻¹

Applying the relation,

ΔT_f = K_f × m

= 12.33 K kg mol⁻¹ × 0.6166 mol kg⁻¹

= 7.60 K (approximately)

Hence, the freezing point of 10% glucose solution is (273.15 − 7.60) K= 265.55 K.