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

9.1 M

Explanation

Weight of glycol = 222.6g
Molar mass of glycol, `C_(2)H_(4)(OH)_(2)`
`=2xx12 +4xx1 +2(16+1)=62"g mol"^(-1)`
Moles of glycol `=(222.6)/(62)=3.59`
Mass of water = 200 g = 0.2 kg
Molality of glycol `=(3.59)/(200) xx 1000=17.95m`
Total mass of solution `=(222.6 +200)g = 422.6g`
Density of solution `= 1.072" g mL"^(-1)`
Volume of solution `=("Mass")/("Density")=(422.6)/(1.072)=394.2mL`
Molarity of solution`=(3.59)/(394.2)xx1000=9.1M`