An antifreeze solution is prepared from 222.6g of ethylene glycol `[(C_(2)H_(6)O_(2))]` and 200g of water (solvent). Calculate the molality of the solution.

Explanation

Weight of Ethylene glycol `= 222.6" gms"`
G.mol wt = 62
Weight of solvent = 200 gms
Molality (m) `=("Wt")/("GMWt")xx(1000)/("a(gms)")`
`=(222.6)/(62)xx(1000)/(200)=(222.6)/(62xx0.2)=17.95" m"`
Molarity (M) `=("Wt")/("GMWt")xx(1000)/("V"(ml))`
`=(222.6)/(62)xx(1000)/(394.2)=9.1" M"`
Mass of solution `=200+222.6`
`=422.6" gms"`
Volume `=("Weight")/("Density")=(422.6)/(1.072)=394.2" ml"`