Question

What mass of ethylene glycol must be added to 5.50 kg of water to lower the freezing point of water from 0°C to −10.0°C?

(K_f for water = 1.86°C kg mol⁻¹, molecular weight of ethylene glycol = 62.0 g mol⁻¹)

Answer 1

Given: W = 5.5 kg = 5500 g

ΔT_f = 10°C

K_f = 1.86 K kg mol⁻¹

m = 62.0 g mol⁻¹

To find: w = ?

Formula: `w = (m xx ΔT_f xx W)/(1000 xx K_f)`

= `(62 xx 10 xx 5500)/(1000 xx 1.86)`

= `(3410000)/(1860)`

= 1833.33 g

 = 18.33 kg