Volumes & Concentrations of Solutions
- The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution
- The solute is the substance that dissolves in a solvent to form a solution
- The solvent is often water
concentration (mol dm-3) =
- A concentrated solution is a solution that has a high concentration of solute
- A dilute solution is a solution with a low concentration of solute
- When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:
- Change mass in grams to moles
- Change cm3 to dm3
- To calculate the mass of a substance present in solution of known concentration and volume:
- Rearrange the concentration equation
number of moles (mol) = concentration (mol dm-3) x volume (dm3)
- Multiply the moles of solute by its molar mass
mass of solute (g) = number of moles (mol) x molar mass (g mol-1)
Worked example
Calculating volume from concentration
Calculate the volume of 1.0 mol dm-3 hydrochloric acid required to completely react with 2.5 g of calcium carbonate.
Answer:
- Write the balanced symbol equation
- CaCO3 + 2HCl → CaCl2 + H2O + CO2
- Calculate the amount, in moles, of calcium carbonate:
- n(CaCO3) = = 0.025 mol
- Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:
- 1 mol of CaCO3 requires 2 mol of HCl
- So 0.025 mol of CaCO3 requires 0.05 mol of HCl
- Calculate the volume of HCl required:
- Volume (HCl) =
- Volume (HCl) = = 0.05 dm3
- So, the volume of hydrochloric acid required is 0.05 dm3
Worked example
Neutralisation calculation
25.0 cm3 of 0.050 mol dm-3 sodium carbonate solution was completely neutralised by 20.0 cm3 of dilute hydrochloric acid.
Calculate the concentration, in mol dm-3, of the hydrochloric acid.
Answer:
- Write the balanced symbol equation:
- Na2CO3 + 2HCl → Na2Cl2 + H2O + CO2
- Calculate the amount, in moles, of sodium carbonate reacted
- n(Na2CO3) = 0.025 dm3 x 0.050 mol dm-3
- n(Na2CO3) = 0.00125 mol
- Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:
- 1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2
- Therefore, 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl
- Calculate the concentration, in mol dm-3 of hydrochloric acid:
- [HCl] =
- [HCl] = = 0.125 mol dm-3
Volumes of gases
- Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’ (also called Avogadro’s hypothesis)
- At room temperature and pressure, one mole of any gas has a volume of 24.0 dm3
- Room temperature is 20 oC
- Room pressure is 1 atmosphere
- Using the following equations, the molar gas volume, 24.0 dm3, can be used to find:
- The volume of a given mass or number of moles of gas
- The mass or number of moles of a given volume of gas
volume of gas (dm3) = amount of gas (mol) x 24.0
amount of gas (mol) =
Worked example
Calculating the volume of gas
Complete the table to calculate the volume that the gases occupy:
Gas | Amount of gas (mol) | Volume of gas (dm3) |
Hydrogen | 3.0 | |
Carbon dioxide | 0.25 | |
Oxygen | 5.4 | |
Ammonia | 0.02 |
Answers:
Gas | Amount of gas (mol) | Volume of gas (dm3) |
Hydrogen | 3.0 | 3.0 x 24.0 = 72.0 |
Carbon dioxide | 0.25 | 0.25 x 24.0 = 6.0 |
Oxygen | 5.4 | 5.4 x 24.0 = 129.6 |
Ammonia | 0.02 | 0.02 x 24.0 = 0.48 |
Worked example
Calculating the number of moles of gas
Complete the table to calculate the number of moles of gas:
Gas | Amount of gas (mol) | Volume of gas (dm3) |
Methane | 225.6 | |
Carbon monoxide | 7.2 | |
Sulfur dioxide | 960 |
Answers:
Gas | Amount of gas (mol) | Volume of gas (dm3) |
Methane | 225.6 / 24.0 = 9.43.0 | 225.6 |
Carbon monoxide | 7.2 / 24.0 = 0.30 | 7.2 |
Sulfur dioxide | 960 / 24.0 = 40 | 960 |