Question

Find the molality of a water solution which freezes at 263.15 K.
(Given: K_f of water = 1.86 K kg mol⁻¹)

Answer 1

Given: Freezing point of solution (T_f) = 263.15 K

Freezing point of pure water = `T_f^circ` = 273.15 K

ΔT_f = 273 − 263.15 = 9.85 K

K_f = 1.86 K kg mol⁻¹

`Delta T_f = K_f * m`

⇒ `m = (Delta T_f)/(K_f)`

= `9.85/1.86`

= 5.296 mol/kg

∴ The molality of the solution is 5.296 mol/kg.