An antifreeze solution is prepared fro, 222.6g of ethylene glycol, `C_2H_4(OH)_2` and 200g 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?

Explanation

Step -1 : Calculation of the molality of the solution.
Mass of ethylene glycol = 222.6 g
Moals mass of ethylene glycol = 62 g `"mol"^(-1)`
Mass of water = 200 g = 0.2 kg
`"Molality of solution (m)"=((222.6g)//(62gmol^(-1)))/(0.2kg)=17.95"mol kg"^(-1)=17.95m`
Step -2 : Calculation of molarity of the solution.
Total mass of solution = Mass of solute + Mass of solvent = 222.6 + 200 = 422.6 g
Density of solution = `.072"g mL"^(-1)`
`"Volume of solution"=("Mass of solution")/("Density solution")=((422.6g))/((1.072gmL^(-1)))=394.2ml=0.3942L`
`"Molarity os solution (M)"=("Mass of ethylene glycol/Molar mass")/("Volume of solution in litres")`
`=((222.6g)//(62gmol^(-1)))/((0.3942L))=9.10" mol L"^(-1)=9.10M`