3.11 Concentrations in mol/dm³ and titration calculations
5 exam questions worth 10 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
0 / 10
How many moles of solute are in 250 cm3 of a 0.20 mol/dm3 solution?
Choose one answer.
Correct Not quite: the answer is 0.050 mol
Mark scheme
B
Model answer
B: 0.20 × 0.250 = 0.050 mol.
A solution of hydrochloric acid has a concentration of 0.50 mol/dm3.
Calculate its concentration in g/dm3. (M_\mathrm{r} of HCl = 36.5)
Mark scheme
- 0.50 × 36.5 = 18.25 g/dm3 (accept 18.3) (1)
Model answer
0.50 × 36.5 = 18.25 g/dm3
25.0 cm3 of potassium hydroxide solution is neutralised by 22.5 cm3 of 0.200 mol/dm3 hydrochloric acid.
KOH + HCl → KCl + H2O
Calculate the concentration of the potassium hydroxide in mol/dm3.
Mark scheme
- moles HCl = 0.200 × 0.0225 = 0.00450 (1)
- moles KOH = 0.00450 (1:1) (1)
- 0.00450 ÷ 0.0250 = 0.180 mol/dm3 (1)
Model answer
Moles of HCl = 0.00450, so moles of KOH = 0.00450. Concentration = 0.00450 ÷ 0.0250 = 0.180 mol/dm3.
20.0 cm3 of 0.150 mol/dm3 sulfuric acid neutralises 25.0 cm3 of sodium hydroxide solution.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Calculate the concentration of the sodium hydroxide in mol/dm3.
Mark scheme
- moles H2SO4 = 0.150 × 0.0200 = 0.00300 (1)
- moles NaOH = 2 × 0.00300 = 0.00600 (1)
- 0.00600 ÷ 0.0250 = 0.240 mol/dm3 (1)
Model answer
Moles of acid = 0.00300; the ratio is 1 : 2 so moles of NaOH = 0.00600. Concentration = 0.00600 ÷ 0.0250 = 0.240 mol/dm3.
What volume of 0.100 mol/dm3 hydrochloric acid, in cm3, is needed to react with 0.00250 mol of sodium hydroxide?
NaOH + HCl → NaCl + H2O
Mark scheme
- moles HCl = 0.00250 (1)
- volume = 0.00250 ÷ 0.100 = 0.0250 dm3 = 25.0 cm3 (1)
Model answer
0.00250 mol of HCl is needed; volume = 0.00250 ÷ 0.100 = 0.0250 dm3 = 25.0 cm3.
Found a mistake?