**1. **Calculate the concentration of the solution when 0.055 mol are dissolved in 25 cm^{3}.

*Answer:* 2.2 mol/dm^{3}

**2. **Calculate the number of moles of solute dissolved in 27.9 cm^{3} of a 0.780 mol/dm^{3} solution.

*Answer:* 0.0218 mol

**3. **What is the concentration of a 0.45 g/dm^{3} solution of copper sulfate in mol/dm^{3} ?

Relative formula mass (M_{r}) = 159.5

*Answer:* 0.00282 mol/dm^{3}

**4. **A student prepared a solution of potassium hydroxide (KOH) by dissolving 10.0 g of solid KOH into 250 cm^{3} of water.

a) Calculate the concentration in g/dm^{3} of the potassium hydroxide solution prepared by the student.

*Answer:* Concentration = mass ÷ volume in dm^{3} ^{3} = 40.0 g/dm^{3}

b) Describe how the concentration of this solution will change if the volume of water used is doubled to 500 cm^{3}.

*Answer:* The concentration will halve to 20.0 g/dm^{3} . Doubling the volume halves the concentration.

c) Another student titrated 25.0 cm^{3} of dilute nitric acid (HNO_{3} ) with a 0.110 mol/dm^{3} solution of potassium hydroxide. 18.6 cm^{3} of the potassium hydroxide solution reacted. The equation for the reaction is: _{3} + KOH → KNO_{3} + H_{2}O

Calculate the concentration of the dilute nitric acid.

*Answer:* mol KOH = concentration × volume ^{3} × (18.6 ÷ 1000) = 0.00205 mol_{3} = mol KOH because of 1:1 ratio in balanced equation ; _{3} = mol ÷ volume^{3}

d) Calculate the mass of potassium hydroxide dissolved in 18.6 cm^{3} of the potassium hydroxide solution. Relative formula mass (M_{r}) KOH = 56

*Answer:* mol KOH = concentration × volume ^{3} × (18.6 ÷ 1000) = 0.00205 mol_{r} = 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 cm^{3} of oxalic acid solution with a 0.225 mol/dm^{3} solution of sodium hydroxide. The equation for the reaction is: _{2}O_{4}H_{2} + 2NaOH → Na_{2}C_{2}O_{4} + 2H_{2}O

31.6 cm^{3} of the sodium hydroxide solution reacted with the oxalic acid solution.

a) Calculate the concentration of the oxalic acid in mol/dm^{3}.

*Answer:* mol NaOH = concentration × volume ^{3} × (31.6 ÷ 1000) = 0.00711 mol

from the ratio in balanced equation, _{2}O_{4}H_{2} = ^{1}⁄_{2} × mol NaOH = 0.00356 mol

concentration C_{2}O_{4}H_{2} = mol ÷ volume ^{3}

b) Use the answer above to calculate the mass of oxalic acid dissolved in 25.0 cm^{3} of the oxalic acid solution. Relative formula mass (M_{r}) C_{2}O_{4}H_{2} = 90

*Answer:* mol C_{2}O_{4}H_{2} = concentration × volume = 0.142 mol/dm^{3} × (25.0 ÷ 1000) = 0.00356 mol ;

mass C_{2}O_{4}H_{2} = moles × M_{r} = 0.00356 × 90 = 0.320 g