Question

An antifreeze solution is prepared from `222.6 g` of ethylene glycol `[C_(2)H_(4)(OH)_(2)]` and `200 g` of water. Calculate the molality of the solution. If the density of the solution is `1.072g mL^(-1)` then what shall be the molarity of the solution?

Answer 1

Molality of ethylene glycol `=(222.6)/(62xx(200)/(1000))`
`= 17.95 m`
`"Wt. of solution"="wt. of gylcol"+"wt. of water"`
`222.6+200= 422.6 g`
Volume of solution `=(422.6)/(1.072)mL`
molarity of ethylene glycol `=(222.6)/(62xx(422.6)/(1.072xx1000))`
`= 9.11 M`