Using Concentrations of Solutions

1. Calculate the concentration of the solution when 0.055 mol are dissolved in 25 cm3.

Answer: 2.2 mol/dm3


2. Calculate the number of moles of solute dissolved in 27.9 cm3 of a 0.780 mol/dm3 solution.

Answer: 0.0218 mol


3. What is the concentration of a 0.45 g/dm3 solution of copper sulfate in mol/dm3 ?
Relative formula mass (Mr) = 159.5

Answer: 0.00282 mol/dm3


4. A student prepared a solution of potassium hydroxide (KOH) by dissolving 10.0 g of solid KOH into 250 cm3 of water.

a) Calculate the concentration in g/dm3 of the potassium hydroxide solution prepared by the student.

Answer: Concentration = mass ÷ volume in dm3 = 10.0 g ÷ 0.250 dm3 = 40.0 g/dm3


b) Describe how the concentration of this solution will change if the volume of water used is doubled to 500 cm3.

Answer: The concentration will halve to 20.0 g/dm3 . Doubling the volume halves the concentration.


c) Another student titrated 25.0 cm3 of dilute nitric acid (HNO3 ) with a 0.110 mol/dm3 solution of potassium hydroxide. 18.6 cm3 of the potassium hydroxide solution reacted. The equation for the reaction is: HNO3 + KOH  →  KNO3 + H2O
Calculate the concentration of the dilute nitric acid.


Answer: mol KOH = concentration × volume = 0.110 mol/dm3 × (18.6 ÷ 1000) = 0.00205 mol ; mol HNO3 = mol KOH because of 1:1 ratio in balanced equation ; concentration HNO3 = mol ÷ volume = 0.00205 ÷ (25.0 ÷ 1000) = 0.08184 mol/dm3


d) Calculate the mass of potassium hydroxide dissolved in 18.6 cm3 of the potassium hydroxide solution. Relative formula mass (Mr) KOH = 56

Answer: mol KOH = concentration × volume = 0.110 mol/dm3 × (18.6 ÷ 1000) = 0.00205 mol ; mass KOH = moles × Mr = 0.00205 × 56 = 0.115


5. Oxalic acid is a naturally occurring acid found in many foods. A student carried out a titration in order to determine the concentration of oxalic acid in a solution. They titrated 25.0 cm3 of oxalic acid solution with a 0.225 mol/dm3 solution of sodium hydroxide. The equation for the reaction is: C2O4H2 + 2NaOH  →  Na2C2O4 + 2H2O
31.6 cm3 of the sodium hydroxide solution reacted with the oxalic acid solution.

a) Calculate the concentration of the oxalic acid in mol/dm3.

Answer: mol NaOH = concentration × volume = 0.225 mol/dm3 × (31.6 ÷ 1000) = 0.00711 mol ;
from the ratio in balanced equation, mol C2O4H2 = 12 × mol NaOH = 0.00356 mol ;
concentration C2O4H2 = mol ÷ volume = 0.00356 ÷ (25.0 ÷ 1000) = 0.1422 mol/dm3


b) Use the answer above to calculate the mass of oxalic acid dissolved in 25.0 cm3 of the oxalic acid solution. Relative formula mass (Mr) C2O4H2 = 90

Answer: mol C2O4H2 = concentration × volume = 0.142 mol/dm3 × (25.0 ÷ 1000) = 0.00356 mol ;
mass C2O4H2 = moles × Mr = 0.00356 × 90 = 0.320 g