Question

A solution containing 54 g of glucose (molecular mass = 180 g mol⁻¹) in 250 g of water (K_f for water = 1.86 K kg mol⁻¹).

What will be the freezing point of this glucose solution?

Options

• 276.402 K

• 270.768 K

• 370.402 K

• 272.563 K

Answer 1

270.768 K

Explanation:

Given: Mass of glucose = 54 g

Molar mass of glucose = 180 g/mol

Mass of water = 250 g = 0.250 kg

K_f = 1.86 K kg mol⁻¹

Freezing point of pure water = 273.0 K

Moles of glucose (n) = `54/180` = 0.3 mol

Molality (m) = `0.3/0.250` = 1.2 mol/kg

ΔT_f ​= K_f​ m

= 1.86 × 1.2

= 2.232 K

`T_f​ = T_"pure" ​- Delta T_f​`

= 273.0 − 2.232

= 270.768 K