Page 395 - SUBSEC October 2017_Neat

02112020/CAPE/SPEC/MS/2017
- 2 –

CHEMISTRY
UNIT 01 — PAPER 02
MARK SCHEME

Question 1

S.O.: Module: 1 ----3.1, 3.2, 3.3, 3.5, 3.6, 3.8, 3.9, 6.1, 6.2, 6.4, 6.6, 6.7,
6.16

KC UK XS
(a) (i) mole: an amount of a substance that contains as many
particles as C atoms in 12g of C-12 isotope [2 marks] 2

[C amount of a substance containing 6 x 10 particles –
23
1 mark only]

(ii) Mass of 1 mole of a substance in grams [1 mark]
1
(iii) Unit of molar mass is g mol -1 [1 mark]
1
(iv) Avogadro’s Law: Equal volumes of all gases contain the
same number of molecules under the same conditions of 1
temperature and pressure. [1 mark]

(b) (i) Molar mass H3PO4 = 3 + 31 + 64 = 98g
98g of H3PO4 is the mass of 1 mole
1.96g of H3PO4 is the 1/98 x 1.96 mole = 0.02 mole
[1 mark]
3
1000 cm H3PO4 soln contains 0.02 mole 2
.
∴20 cm H3PO4 soln contains X 20
3

= 0.0004 mole [1 mark]

(ii) No of mole NaOH = X 1.28 = 0.032 mole

.
3
25 cm Na OH solution contains X 25 1

= 0.0008 mole [1 mark]

(iii) 0.0004 mole acid reacts with 0.0008 mole NaOH
.
∴1 mole acid reacts with
.
= 2 moles [1 mark] 1

(iv) H3PO4(aq) + 2 NaOH(aq) Na2HPO4(aq)+2H2O(l) [1 mark] 1

(v)
 Measure 25 cm NaOH solution with a clean pipette
3
3
 Deliver this solution in a clean 250 cm conical flask

 Add a few drops of phenolphthalein indicator
 Fill a clean burette to the zero mark with phosphoric
acid 5
 Add acid dropwise to the alkali until the solution goes
colourless [5 marks]