Answer:
17.54 g
Explanation:
the freezing point of the solution to -8.8°C
depression in freezing point = 8.8°C
Molal freezing point depression constant of ethylene glycol Kf = 3.11⁰C /m
ΔTf = Kf x m , m is no of moles of solute per kg of solvent .
Let gram of ethylene glycol required be x .
moles = m / mol weight of ethylene glycol
= x / 62
100 g of water = .1 kg
moles of solute dissolved in 1 kg of water
m = x / 62 x .1
m = 10 x / 62
Using the above equation for depression in freezing point
8.8 = 3.11 x 10 x / 62
x = 17.54 g .
