Converting molality to molarity: Ethylene glycol, `C_(2)H_(4)(OH)_(2)` is a colourless liquid used as an automobile antifreeze. If the density at `20^(@)C` of a `4.500 m` solution of ethylene glycon in water is `1.279 g mL^(-1)`, what is the molarity of the solution? The molar mass of ethylene giycon is `62.00 g mol^(-1)`.
Strategy: A `4.500 m` solution of ethylene glycol water contains `4.500 mol` of glycol in every one kilogram of water. To find the solution's molarity, we must find the number of moles of solute per litre of solution. We can find volume fome the mass the mass of the solution by using density as a conversion factor.
So we cam take the mass of solution that contains `1 kg` of `H_(2)O` and use the density to convert this quantity of solution to volume.

Explanation

Step 1. Finding the mass of glycol The mass of the solution is the sum of the masses of solute and solvent. Assusming that `1.000 kg` of `H_(2)O` is used to dissolve `4.500 mol` of glycol, the mass of the glycol is
`n_(glycol)=(mass)_(glycol)/("molar mass"_(glycol))`
or `"mass"_(glycol)=(n_(glycol))xx("molar mass"_(glycol))`
`=(4.500 mol)(62.00 g//mol)`
`=279.0 g`
Step 2. Finding the mass of solution. Dissolving this `279.0 g` of ethylene glycol in `1.000 kg (or 1000 g)` of water gives the total mass of the solution:
Mass of solution= mass of solute + mass of solvent
`=279.0 g+ 1000 g`
`=1279 g`
Step 3. Finding the volume of solution
The volume of the solution is obtained from its mass by using density as a conversion factor:
`("density")_("soln")=(("mass")_("soln"))/(("volume")_("soln"))`
or `("volume")_("soln")=(("mass")_("soln"))/(("density")_("soln"))`
`=(1279 g)/(1.279 g//mL)`
`=1000 mL`
`=1.000 L`
Step 4. Finding the Molarity of solution. The molarity of the solution is the number of moles of solute divided by the volume of solution.
Molarity of solution=`n_(glycol)/V_(L)`
`=(4.500 mol)/(1.000 L)`
`=4.500 M`