Concentration Calculations (Edexcel International A Level (IAL) Chemistry): Revision Note

Mass concentration calculations

1. What is the mass concentration when 6.34 g of sodium chloride is dissolved into a 0.250 dm³ solution?

2. A sodium carbonate solution has a mass concentration of 5.2 g dm^-3. What is the volume of the solution made when 250 g of sodium carbonate is used?

3. The mass concentration of a solution is 26.7 g dm^-3. What is the mass of sodium bromide in 500 cm³ of solution?

Answer 1

• Mass concentration $=\frac{\mathrm{mass}\left(\mathrm{g}\right)}{\mathrm{volume}\left({\mathrm{dm}}^3\right)}=\frac{6.34}{0.250}=$25.4 g dm^-3 (to 3 s.f.)

   Answer 2

• Volume of solution $=\frac{\mathrm{mass}\left(\mathrm{g}\right)}{\mathrm{mass}\mathrm{concentration}\left(\mathrm{g}{\mathrm{dm}}^3\right)}=\frac{250}{5.2}=$48 dm³

  • This answer should be given to 2 s.f. as there is a value in the question with only 2 significant figures

   Answer 3

• Mass = mass concentration (g dm^-3) x volume of solution (dm³) = 26.7 x 0.5 = 13.4 g (to 3 s.f.)  

  • The 500 cm³ in the question has to be converted into dm³ 

  • $\frac{500}{1000}$ = 0.5 dm³ 

Molar concentration calculations

1. What is the molar concentration when 6.34 g of sodium chloride is dissolved into a 250 cm³ solution?

2. A sodium carbonate solution has a molar concentration of 1.25 mol dm^-3. What is the volume of the solution made when 250 g of sodium carbonate is used?

3. The molar concentration of a sodium bromide solution is 0.250 mol dm^-3. What is the mass of sodium bromide in 500 cm³ of this solution?

Answer 1

• Number of moles of NaCl $=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{6.34}{(23.0+35.5)}=$0.1084 moles

  • Molar concentration in mol dm^-3 $=\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\left(\mathrm{or}\mathrm{amount}\right)}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}{\mathrm{dm}}^3}=\frac{0.1084}{0.250}=$0.434 mol dm^-3 

   Answer 2

• Number of moles of Na₂CO₃ $=\frac{\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{250}{(23.0\times 2)+12.0+(16.0\times 3)}=$2.358 moles

  • Volume of solution in dm³

$=\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\left(\mathrm{or}\mathrm{amount}\right)}{\mathrm{Molar}\mathrm{concentration}\mathrm{in}\mathrm{mol}{\mathrm{dm}}^{-3}}=\frac{2.358}{1.25}=$1.89 dm³ 

   Answer 3

• Number of moles of NaBr = molar concentration x volume of solution

= 0.250 x 0.500 = 0.125 moles 

• Mass of NaBr = number of moles of NaBr x molar mass

= 0.125 x (23.0 + 79.9) = 12.9 g

The concentration of chlorine in a swimming pool should between between 1 and 3 ppm. Calculate the maximum mass, in kg, of chlorine that should be present in an olympic swimming pool of size 2.5 million litres.

Answer:

Step 1: calculate the total mass in mg assuming 3ppm(1 litre is the same as 1 dm³)

• 3 x 2.5 x 10⁶ = 7.5 x 10⁶ mg

Step 2: convert the mass into kilograms (1 mg = 10^-6 kg)

• 7.5 x 10⁶  x 10^-6  kg = 7.5 kg

Atmospheric gas concentration calculations

Calculate the concentration, in ppm, of the following:

1. A volume of 2.5 dm³ of carbon dioxide in 10000 dm³ of air

2. A volume of 2.5 dm³ of sulfur dioxide in 4000 dm³ of air

3. A volume of 152 cm³ of ozone in 112 dm³ of air

  Answer 1

• Concentration in ppm

$=\frac{\mathrm{volume}\mathrm{of}\mathrm{gas}\times 1000000}{\mathrm{volume}\mathrm{of}\mathrm{air}}=\frac{2.5\times 1000000}{10000}=$ 250 ppm 

   Answer 2

• Concentration in ppm

$=\frac{\mathrm{volume}\mathrm{of}\mathrm{gas}\times 1000000}{\mathrm{volume}\mathrm{of}\mathrm{air}}=\frac{2.5\times 1000000}{4000}=$ 625 ppm 

   Answer 3

A volume of 152 cm³ of ozone in 112 dm³ of air

• Concentration in ppm $=\frac{\mathrm{volume}\mathrm{of}\mathrm{gas}\times 1000000}{\mathrm{volume}\mathrm{of}\mathrm{air}}=\frac{0.152\times 1000000}{112.0}=$1360 ppm 

When completing atmospheric gas calculations, the gas involved does not affect the calculation as shown by worked examples 1 and 2