Reacting Volumes | AQA AS Chemistry Revision Notes 2016

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) = $\frac{number of moles of solute (mol)}{volume of solution ({dm}^{3})}$

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^-3the following points need to be considered:
- Change mass in grams to moles
- Change cm³to 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)

mass of solute (g) = number of moles (mol) x molar mass (g mol^-1)

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) = $\frac{2.5 g}{100 g {mol}^{- 1}}$ = 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) = $\frac{amount (mol)}{concentration (mol {dm}^{- 3})}$
- Volume (HCl) = $\frac{0.05 mol}{1.0 mol {dm}^{- 3}}$ = 0.05 dm³
- So, the volume of hydrochloric acid required is 0.05 dm³

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] = $\frac{amount (mol)}{volume ({dm}^{3})}$
- [HCl] = $\frac{0.00250}{0.0200}$ = 0.125 mol dm^-3

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) = $\frac{volume of gas ({dm}^{3})}{24.0}$

Calculating the volume of gas

Complete the table to calculate the volume that the gases occupy:

Answers:

Calculating the number of moles of gas

Complete the table to calculate the number of moles of gas:

Answers: