To change a percentage by mass into a molar concentration, it is easiest to consider 1.00 L of the solution. Its density gives its mass, the percentage gives the mass of solute, and the molar mass gives the amount.
Concentrated sulfuric acid is 98.0 % sulfuric acid by mass and has a density of 1.84 g/cm³. So 1.00 L, which is 1000 cm³, has a mass of 1840 g, of which 1840 g × 0.980 = 1803 g is sulfuric acid, \mathrm{H_2SO_4}.
With a molar mass of 98.1 g/mol, this is 1803 g ÷ 98.1 g/mol = 18.4 mol, so the concentration is 18.4 mol/L. When it is diluted, the acid is added slowly to water, never water to the acid, because much heat is given out.
The reverse calculation also starts from 1.00 L. A 1.00 mol/L sodium hydroxide solution has a density of about 1.04 g/cm³, so 1.00 L of it has a mass of 1000 cm³ × 1.04 g/cm³ = 1040 g.
This litre contains 1.00 mol of sodium hydroxide, \mathrm{NaOH}, which is 1.00 mol × 40.0 g/mol = 40.0 g. The percentage by mass is 40.0 g ÷ 1040 g × 100 = 3.85 %.
A molar concentration alone does not give the percentage by mass, or the other way round. The density of the solution is always needed to connect a volume with a mass.
Adding water to a solution lowers its concentration but does not change the amount of solute, because no solute is added or taken away.
If 10.0 mL of 1.00 mol/L solution, which holds 1.00 mol/L × 0.0100 L = 0.0100 mol of solute, is made up to 100 mL with water, the same 0.0100 mol is now in 0.100 L, so the concentration is 0.100 mol/L.
Making the volume ten times larger makes the concentration ten times smaller.
Because the amount of solute stays the same, the concentration and volume before dilution, c_1 and V_1, and after dilution, c_2 and V_2, are related by c_1V_1 = c_2V_2.
To make 100 mL of 0.100 mol/L solution from 1.00 mol/L solution, the volume needed is 0.100 mol/L × 100 mL ÷ 1.00 mol/L = 10.0 mL. Both volumes can stay in millilitres here, because the units cancel.
The 10.0 mL is measured with a 10 mL volumetric pipette and run into a 100 mL volumetric flask. Water is then added to the mark and the solution is mixed.
When two solutions of the same solute are mixed, the amounts of solute add up. A mixture of 100 mL of 0.200 mol/L solution, with 0.0200 mol, and 300 mL of 0.100 mol/L solution, with 0.0300 mol, contains 0.0500 mol.
For dilute solutions the volumes can be taken to add up too, giving 400 mL, so the concentration is 0.0500 mol ÷ 0.400 L = 0.125 mol/L, between the two starting concentrations.
For concentrated solutions, such as sulfuric acid and water, the volumes do not add up exactly, and the volume of the mixture has to be measured.