Converting molarity to mole fraction, mass percent and molality: A `0.750 M` solution of `H_(2)SO_(4)` in water has a density of `1.049 g mL^(-1)` at `20 ^(@)C`. What is the concentration of this solution in (a) mole fraction, (b) mass percent. And (c ) molality?
Strategy: To solve this problem, we need to know the mass of solute as well as mass of solvent. There is `0.750 mol H_(2)SO_(4)` per liter of solution. Let's pick an arbitrary amount of the solution that will make the calculations easy, say `1.00 L`. Consider `1 L (=1000 ml)` of solution and calculate its mass, using the density. We can then calculate the mass of `H_(2)SO_(4)` and find the mass of `H_(2)O` by difference. Now we can calculate the mole fraction, mass precent and morality.

Explanation

Step 1. Find the mass of solute since the concentration of the solution is `0.75 mol L^(-1), 1.00 L` of the solution contains 0.750 mol of `H_(2)SO_(4)`.
`n_(H_(2)SO_(4))=mass_(H_(2)SO_(4))/("molar mass"_(H_(2)SO_(4)))`
or `mass_(H_(2)SO_(4))=n_(H_(2)SO_(4))xx"molar mass"_(H_(2)SO_(4))`
`=(0.750 mol)(98.0 g mol^(-1))`
`=73.5 g`
Step 2. Finding the mass of solution. Since the density of the solution is `1.046 g mL^(-1), 1.00 L` of the solution has a mass of `1.049 kg`:
`("density")_("slon")=(("mass")_("soln"))/(("volume")_("soln"))`
`or ("mass")_("soln")=("volume")_("soln") ("density")_("soln")`
`=(1.00 L)(1.049 Kg L^(-1))`
`=1.049 kg`
`=1049 g`
Step 3. Finding the mass of water (solvent) and its number of moles Subtracting the mass of solute `(H_(2)SO_(4))` from the total mass of the solution gives mass of solvent:
Mass of `H_(2)O` (solvent) in 1.00 L of solution
`=(mass)_(solution)-(mass)_(solution)`
`=(1049 g)-(73.5 g)`
`=975.5 g`
Dividing the mass of `H_(2)O` by its molar mass gives number of moles of water:
`n_(H_(2)O)=((mass)_(H_(2)O))/((molar mass)_(H_(2)O))`
`=(975.5 g)/(18 g mol^(-1))`
`=54.2 mol H_(2)O`
Step 4. Finding the mole fraction of `H_(2)SO_(4)`
`chi_(H_(2)SO_(4))=n_(H_(2)SO_(4))/(n_(H_(2)SO_(4))+n_(H_(2)O))`
`=(0.750 mol H_(2)SO_(4))/(0.750 mol H_(2)SO_(4) + 54.2 mol H_(2)O)`
`=0.0136`
Step 5. Finding the mass percent of `H_(2)SO_(4)`
The mass percent concentration can be determined from the calculations in step (1) and (2),
`Mass% H_(2)SO_(4)=((Mass)_(H_(2)SO_(4)))/((Mass)_(soln))xx100%`
`=(73.5 g)/(1049 g)xx100%`
`=7.01 %`
Step 6. Finding the molality of solute `(H_(2)SO_(4))`:
The molality of the solution can be determined from step (1) and step (3), Since `975.5 g` of `H_(2)O` has `0.750` mol of `H_(2)SO_(4)` dissolved in it, we have
Molality (m)=`n_(solute)/g_(solvent)xx(1000 g)/(kg)`
`=(0.750 mol H_(2)SO_(4))/(975.5 g)xx(1000 g)/(kg)`
`=0.769 mol kg^(-1)`
`=0.769 m H_(2)SO_(4)`