(i)

Saponification is a process in which fat/oil/lipid is reacted with alkaline solution to form soap and glycerol.

(ii)

Observation that could be made when brine is added to the products of saponification

Formation of precipitate (of soap is observed).

(i)

A solvent is a substance/liquid/gas/solid that dissolves other materials to form a solution.

(ii)

Note: volume is in dm³.

500 cm³ =

Molar/Concentration = $\frac{500}{1000}$ dm³ = 0.5 dm³

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

Mass = 20 g

Concentration = $\frac{20}{\mathrm{40\; x\; 0.5}}$ = 1.0 mol/dm³

(i)

(ii)

Reaction involving ammonium chloride and calcium hydroxide

2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2H₂ + 2NH₃

(i)

Temporary hard water is the hardness of water caused by the presence of hydrogen carbonates (bicarbonates) of calcium and magnesium and can be softened by boiling/heating.

(ii)

Chemical equation to show how the temporary hard water can be softened by heating

Ca(HCO₃)₂(aq)→CO₂(g)+H₂O(l)+CaCO₃(s)

OR

Mg(HCO₃)₂(aq)→CO₂(g)+H₂O(l)+MgCO₃(s)

(i)

Method of preparation of a salts:

(ii)

Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃

(i)

Apparatuses needed in the preparation of a standard solution

1. Volumetric flask
2. Beaker
3. Funnel
4. Stirrer
5. Chemical balance
6. Spatula
7. Wash bottle
8. Dropper

(ii)

Concentration/Molarity = $\frac{\mathrm{Mole}}{\mathrm{Volume}}$

Mole = Concentration/Molarity x Volume.

The value of the mole is constant even after dilution.

M₁ x V₁ = M₂ x V₂

V₂ = $\frac{{\mathrm{M}}_{1}x{\mathrm{V}}_1}{{\mathrm{M}}_2}$

M₁ = 5 M

V₁ = 20 cm³

M₂ = 1 M

V₂ = $\frac{\mathrm{5\; M}x20{\mathrm{cm}}^3}{\mathrm{1\; M}}$ = 100 cm³

Volume of water needed = V₂ - V₁

Volume of water needed = 100 cm³ - 20 cm³

Volume of water needed = 80 cm³

(i)

Cu + 2H₂SO₄ → CuSO₄ + 2H₂O + SO₂

(ii)

Concentration of H₂SO₄ = 0.5 mol dm^-3

Volume of H₂SO₄ = 25 cm³

Change the cm³ to dm³ by dividing by 1000

25 cm³ = $\frac{25}{1000}$ dm³ = 0.025 dm³

Mole of H₂SO₄ = 0.5 mol dm^-3 x 0.025 dm³

Mole of H₂SO₄ = 0.0125 moles

Using the mole ratios

From the equation, 2 moles of H₂SO₄ reacted with 1 mole of Cu

0.0125 moles of H₂SO₄ will react with $\frac{\mathrm{0.0125\; moles\; x\; 1\; mole}}{2}$ = 0.00625 moles of Cu

Mass of Cu = 0.04 g

Mole of Cu = 0.00625 mol

Molar Mass of Cu = $\frac{\mathrm{0.04\; g}}{\mathrm{0.00625\; mol}}$ = 6.4 gmol^-1 or 6.4 g/mol

(i)

Completed equations

(α) H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
(β) CH₃COOH + CH₃OH → CH₃COOCH₃ + H₂O

(ii)

Names of reactions

(a) Neutralization
(b) Esterification