The concentration equation
Everything on the last page can be done in one line with one equation. Four problems, each solved with it.
Concentration
C = n ÷ V
n in moles (mol), V in cubic decimetres (dm3), C in mol dm-3
Example 1 — sodium nitrate
Problem
17 g of sodium nitrate (NaNO₃) is weighed out accurately, dissolved in a volumetric flask marked at 200 cm³, and made up to a final volume of 200 cm³ with distilled water. What is the concentration of NaNO₃ in the solution? (Na = 23, N = 14, O = 16)
Solution
molar mass of NaNO3 = {23 + 14 + (16 × 3)} g mol-1 = 85 g mol-1
amount in 17 g = 17 g ÷ 85 g mol-1 = 0.2 mol
amount in 1000 cm3 = (0.2 mol ÷ 200 cm3) × 1000 cm3 = 1 mol
concentration = 1 mol ÷ 1 dm3
= 1 mol dm-3
Example 2 — potassium carbonate
Problem
What mass of K₂CO₃ is needed to make 500 cm³ of a potassium carbonate solution of concentration 1 mol dm⁻³? (K = 39, C = 12, O = 16)
Solution
molar mass of K2CO3 = (39 × 2) + 12 + (16 × 3) = 138 g mol-1
mass in 1000 cm3 of a 1 mol dm-3 solution = 138 g
mass in 500 cm3 = (138 g ÷ 1000 cm3) × 500 cm3
= 69 g
Example 3 — urea
Problem
12 g of urea (CO(NH₂)₂) is dissolved in distilled water to prepare 1 dm³ of solution. Find the concentration of this solution. (C = 12, O = 16, N = 14, H = 1)
Solution
molar mass of urea = {12 + 16 + (14 × 2) + (1 × 4)} g mol-1 = 60 g mol-1
amount in 12 g = (1 mol ÷ 60 g) × 12 g = 0.2 mol
concentration = 0.2 mol ÷ 1 dm3
= 0.2 mol dm-3
Example 4 — glucose, with a conversion in it
Problem
18 g of glucose is put into a 250 cm³ volumetric flask and distilled water is added until the solution reaches 250 cm³. Find the concentration of this solution.
Solution
molar mass of glucose = (12 × 6) + (1 × 12) + (16 × 6) = 180 g mol-1
amount in 18 g = (1 mol ÷ 180 g) × 18 g = 0.1 mol
amount in 1000 cm3 = (0.1 mol ÷ 250 cm3) × 1000 cm3 = 0.4 mol
= 0.4 mol dm-3