Specification Point 1.34C (Paper 2C Only):
Understand how to carry out calculations involving Amount of Substance, Volume and Concentration (in mol / dm3 ) of Solution

General Equation:

Substance, concentration, volume equation, Edexcel IGCSE Chemistry

Calculating Moles

Equation:

Amount of Substance (mol)    =   Concentration    x    Volume of Solution (dm3)

Example:

Calculate the Moles of Solute Dissolved in 2 dm3 of a 0.1 mol / dm3 Solution

Concentration of Solution : 0.1 mol / dm3

Volume of Solution : 2 dm3

Moles of Solute   =   0.1   x   2   =   0.2

Amount of Solute = 0.2 mol

Calculating Concentration

Equation:

Substance, concentration, volume equation, Edexcel IGCSE Chemistry

Example:

25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration, in mol / dm3 of the hydrochloric acid.

Step 1 – Calculate the amount, in moles, of sodium carbonate reacted

              Amount of Na2CO3  =  ( 5.0 x 0.050 ) ÷ 1000  =  0.00125 mol

Step 2 – Calculate the amount, in moles, of hydrochloric acid reacted

              Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

              1 mol of Na2CO3 reacts with 2 mol of HCl

              0.00125 mol of  Na2CO3 Reacts with 0.00250 mol of HCl

Step 3 – Calculate the concentration, in mol / dm3, of the Hydrochloric Acid

              1 dm3 = 1000 cm3

              Concentration ( mol / dm3 ) =  0.00250 ÷ ( 20 ÷ 1000 )  =   0.125

                                         Concentration of Hydrochloric Acid = 0.125 mol / dm3

Calculating Volume

Equation:

Equation giving volume from substance:concentration, Edexcel IGCSE Chemistry

Example:

Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm3 that is required to react completely with 2.5g of calcium carbonate.

Step 1 – Calculate the amount, in moles, of calcium carbonate that reacts

              Mr of CaCO3 is 100

               Amount of CaCO3  =  ( 2.5 ÷ 100 )  =  0.025 mol

Step 2 – Calculate the moles of hydrochloric acid required

               CaCO3 +  2HCl  → CaCl2  +  H2O  +  CO2

               1 mol of CaCO3 requires 2 mol of HCl

               0.025 mol of CaCO3 Requires 0.05 mol of HCl

Step 3 – Calculate the volume of HCl Required

Volume  =  ( mol of Substance  ÷  Concentration )

                             =  0.05  ÷  1.0

                             =  0.05 mol

                            Volume of Hydrochloric Acid = 0.05 mol

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Author: Jamie

Jamie got a First class degree in Chemistry from Oxford University before going on to teach chemistry full time as a professional tutor. He’s put together these handy revision notes to match the Edexcel IGCSE Chemistry specification so you can learn exactly what you need to know for your exams.