Converting mass percent to molarity: the density of a 24.5 mass % solution of sulphuric acid `(H_(2)SO_(4))` in water is `1.176 g mL^(-1)` at `25.0^(@)C`. What is the molarity of the solution?
Stregy: Use mass % to calculate number of moles of solute and use density of solution to calculate volume of solution. Describing a solution as 24.5 mass % `H_(2)SO_(4)` in `H_(2)O` means that every 100.0 g of solution contains 24.5 g of `H_(2)SO_(4)` and 75.5 g of `H_(2)O`. Since we want to calculate the concentration in molarity, we first need to find the number of moles of `H_(2)SO_(4)` dissolved in a specific mass of solution. Next we use density as a conversion factor to find the volume of that solution and then calculate molarity by dividing the number of moles of `H_(2)SO_(4)` by the volume of solution.

Explanation

Consider `100.0 g` of solution. Therefore, mass of `H_(2)SO_(4)` is `24.5 g`. First, convert the mass of `H_(2)SO_(4)` into moles.
`n_(H_(2)SO_(4))=(Mass H_(2)SO_(4))/(Molar mass H_(2)SO_(4))`
`=(24.5 g H_(2)SO_(4))/(98.0 g H_(2)SO_(4)//mol H_(2)SO_(4))`
`=0.25 mol H_(2)SO_(4)`
Next, we find the volume of `100.0 g` of solution using sensity as the conversion factor:
Volume of solution`=((1216g)/(1000mL))/((100.0" g soln")/(1.176"g soln"//mL))`
`=85mL`
`=85 mL`
Finally, calculate the molarity of solution using Eqution `2.14`.
Molarity`=n_(H_(2)SO_(4))/V_(mL)xx(1000 mL)/L`
`=(0.25 mol H_(2)SO_(4))/(85 mL soln)xx(1000 mL)/L`
`=2.94 M`
Thus, the molarity of the `24.5` mass `%` of sulphuric acid solution is `2.94 M`.
Short cut method: We can also convert mass percent into molarity by using the following formula:
`M=("(mass%)(specific gravity or density of soln)(10)")/(("molar mass of solute"))`
`=((24.5)(1.175)(10))/98`
`=2.94 mol L^(-1)`