An antifreeze solution is prepared from 222.6 g of ethylene glycol `[C_2H_4 (OH)_2]` and 200 g of water. Calculate the molality of the solution. If the density of this solution be `1.072 g mL^(-1)` , what will be the molarity of the solution ?

Explanation

Applying the following relation and substituting the values, we get
`m = (W_B)/(M_B) xx (1000)/(W_A) = (222.6)/(62) xx 1000/200 = 2226/124 = 17.95`mol/kg
Molality (m) and molarity (M), are related as follows :
`m = (M)/(d - M xx M_B xx 10^(-3))`
` 17.95 = (M)/(1.072 - M xx 62 xx 10^(-3))`
`17.95 xx 1.072 - 17.95 xx 62 xx 10^(-3) M = M`
`M (1 + 17.95 xx 62 xx 10^(-3)) = 19.24 " or " M = (19.24)/(2.11) = 9.12` mol/litre
Molarity of the solution = 9.12 M.