Titration Calculations - 3

Titrations are often used to find the purity of a solid or its M_r. The solid is weighed, dissolved and made up to 250.0 cm³ in a volumetric flask, and 25.0 cm³ portions (aliquots) are titrated. Because each portion is one tenth of the flask, you must scale up by 10 at the end.Worked example: 2.00 g of impure sodium hydroxide was dissolved and made up to 250.0 cm³. 25.0 cm³ portions needed a mean titre of 22.50 cm³ of 0.200 mol dm^-3 hydrochloric acid. Calculate the percentage purity of the sodium hydroxide.NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)Step 1: amount of HCl = 0.200 × (22.50 ÷ 1000) = 4.50 × 10^-3 molStep 2: ratio 1 : 1, so amount of NaOH in 25.0 cm³ = 4.50 × 10^-3 molStep 3: amount of NaOH in 250.0 cm³ = 4.50 × 10^-3 × 10 = 4.50 × 10^-2 molStep 4: mass of NaOH = 0.0450 × 40.0 = 1.80 gStep 5: percentage purity = (1.80 ÷ 2.00) × 100 = 90.0%The same scale-up lets you find an unknown M_r: after step 3, divide the mass of solid dissolved by the total number of moles.Forgetting the factor of 10 is the classic error here, so mark the scale-up as its own step in your working.