Observation

Majority of the candidates responded to the question and their performance was average.

In part (a), few candidates calculated the number of atoms that are in 10.0 g of CaCO3.

In part (b), majority of the candidates could not write a balanced equation to show the acidic nature of C2H2 when reacted with sodium.

In part (c), majority of the candidates could not draw the structures of the compounds X and Y respectively.

In part (d), majority of the candidates could not write the half-reaction equation when magnesium reacts with aqueous solution of an acid to liberate hydrogen gas.

In part (e), majority of the candidates stated the differences between a mixture and a compound.
The expected answers include:

  (a)      Molar Mass of CaCO3 = 40 + 12 + 3 (16)     
= 100 g mol-1

                       :. 10.0 g CaCO3 = 10/100 

                                                 = 0.1 mol

            no. of atoms per molecule = 5
       ⇒ no of atoms in 10.0g CaCO3 = 5 x 0.1 x 6.02 x 10²³
= 3.01 x 1023 atoms
                                                                                                                                   

(b)(i)           I.          C2H2 + 2Na → Na2C2(s) + H2(g) (2)
Or  2C2H2 + 2Na → 2C2HNa + H2
                   II.        2C2H5OH + 2Na → 2C2H5ONa(s) + H2(g)

(ii)      

I.          Sodium Carbide / Sodium ethynide   
II.        Sodium ethoxide

(ii)       Compound X decolourises bromine water but compound Y does not decolourise bromine water.

(iii)      Because compound X is an unsaturated compound and compound Y is a saturated compound

(d)(i)  Mg(s) →  Mg²⁺(aq)+ 2e^-        (1)

           2H⁺(aq) + 2e^- → H2(g)  (1)

      (ii)  Mg(s) + 2H⁺(aq) →  Mg²⁺(aq) + H2(g)(1)
                                                                       
(e)(i)

                                                                                                           
(ii)  quick lime – compound
Gold – element 
Soil – mixture