Question

To make a solution of urea (Molecular mass = 60 gm/mole) having freezing point of 0.372°C How many grams of urea should be added in 8 Kg of water ? (Kf = 1.86o C kg mole1

Answer 1

Answer:96g of urea

Explanation:

ΔT=K*molality

0.372=1.86*(m/60)*(1000/8000)

m=0.372*60*8/1.86

m=96g

I hope this helps you friend!!