Liquids/Solutions: Moles, Volume & Conc.

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:

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

Equation:

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 by  rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm³ to dm³      

      Amount of Na₂CO₃  =  (25.0 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

Equation:

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

              M_r 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  =  Amount of substance (mol)  ÷  Concentration (mol / dm³)

                           =  0.05  ÷  1.0

                           =  0.05 dm³ (the moles cancel out above and below the line)

Volume of Hydrochloric Acid = 0.05 dm³