Question
What approximate proportions by volume of water( d=1g/cc) and ethylene glycol C2H6O2 (d=1.12g/cc) must be mixed to ensure protection of an automobile cooling system to -10 C?
Joshi sir comment
we know that dT = Kfm so m = 10/1.86
m means moles of ethelene glycol in 1000 gm water
so 1000 gm water contains (10/1.86) * 62 gm ethylene glycol
so 1000 cc water contains [(10/1.86)* 62] / 1.12 cc
now calculate