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 1.072g `mL^(-1)`, then what shall be the molarity of the solution?

Answer

`a m=17.95, b. M=9.1`

Explanation

`Mw` of ethylene glycol `=62g`
`a.` `m=(W_(2)xx1000)/(Mw_(2)xxW_(1))=(222.6xx1000)/(62xx200)=17.95mol Kg^(-1)`
`b.` Weight of solution `=222.6+ 200=422.6g`
`d_(sol)=1.072 g mL^(-1)`
Volume of solution `=(422.6)/(1.072)=394.21mL`
`M=(W_(2)xx1000)/(Mw_(2)xxV_(sol))=(222.6xx1000)/(62xx394.21)=9.1M`