Question

Calculate the temperature at which a solution containing 54g of glucose, (C₆H₁₂O₆), in 250g of water will freeze. (Kf for water = 1.86 K mol⁻¹ kg)

Answer 1

Given:

Mass of glucose = 54g

​Mass of water = 250 g

K_f for water = 1.86 K kg mol^-1

​Molecular mass of glucose =  180.15 g/mol

Using the relation:

Now,Freezing point of solution = Freezing point of water − depression in freezing point Freezing point of solution = 273 − 2.23 = 270.77 K

Answer 2

 ΔT_f = K_fm

No. of moles of glucose =

Molality of Glucose solution =

ΔT_f = K_fm

= (1.86 K kg mol^-1) x (1.20 mol kg^-1)

= 2.23 K

Temperature at which solutions freezes = (273.15 - 2.23K)

= 270.77 K or 2.23^°C

or (273.000 - 2.23)K = 270.7 K