Working with concentration

Three worked problems. The first finds a concentration from an amount; the other two go the other way, from a concentration to the mass you have to weigh out.

Example 1 — from moles to a concentration

Problem

If 2 dm³ of a solution contains four moles of sodium hydroxide (NaOH), find the concentration of sodium hydroxide in that solution.

Solution

amount of NaOH in 2 dm³ of the solution = 4 mol

amount of NaOH in 1 dm³ = (4 mol ÷ 2 dm³) × 1 dm³ = 2 mol

concentration of NaOH = 2 mol ÷ 1 dm³

= 2 mol dm^-3

Example 2 — from a concentration to a mass

Problem

What mass of glucose (C₆H₁₂O₆) is needed to make 1 dm³ of a 1 mol dm⁻³ glucose solution? (C = 12, H = 1, O = 16)

Solution

here 1 mol of glucose is needed

molar mass of glucose = {(12 × 6) + (1 × 12) + (16 × 6)} g mol^-1 = 180 g mol^-1

mass of glucose needed = 180 g mol^-1 × 1 mol

= 180 g

Example 3 — the same solution, a smaller volume

Problem

Find the mass of glucose that has to be weighed out to prepare 500 cm³ of a 1 mol dm⁻³ glucose solution.

Solution

mass of glucose needed to make 1000 cm³ = 180 g

mass needed to make 500 cm³ = (180 g ÷ 1000 cm³) × 500 cm³

= 90 g

Half the volume at the same concentration needs half the mass. You can see that without the arithmetic, and it is a good check that the arithmetic agrees.