An antifreeze solution is prepared from 222.6 g of ethylene glycol `(C_(2)H_(6)O_(2))` and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g `mL^(-1)`, then what shall be the molarity of the solution ?

Explanation

Molar mass of ethylene glycol `[C_(2)H_(4)(OH)_(2)]`
`= 2xx12+6xx1+2xx16`
= 62 g `mol^(-1)`
Number of moles of ethylene glycol
`= (222.6g)/(62 g mol^(-1))`
= 3.59 mol
Therefore, molality of the solution
`= (3.59 mol)/(0.20 kg)`
= 17.95 m
Total mass of the solution `= (222.6+200)g`
= 422.6 g
Given,
Density of the solution `=(422.6 g)/(1.072 g mL^(-1))`
= 394.2 mL
`= 0.3942xx10^(-3)L`
Molarity of the solution `= (3.59 mol)/(0.39422xx10^(-3)L)`
= 9.11 M.