In part (a) candidates correctly tabulated their burette readings and calculated the average titre value. Some candidates however manipulated their readings to agree with those of their teachers.

In part (b) (i) and (ii), most candidates that lost marks in this section did so for either using wrong units or not expressing their answers to three significant figures. In section (iii), many candidates correctly calculated the molar mass of Na2CO3 and obtained the mass of anhydrous salt. In the calculation of the value of y however, many candidates failed to express their final answer to the nearest whole number while others muddled up the substitution . Performance was on the average.

The expected answers are:

(b)(i) CDVD = 2 mole ratio

CEVE 1

CE = CDVD making CE subj

2VE

= X moldm-3

(b(i)Amount of D used = __0.100 x V____D__ = d mol

1000

From the balanced equation of the reaction

2 moles of D __=__ 1 mole of E

D mol of D __=__ __ ____d __mole of E

2

i.e 20/25 cm3 of E contains d/2 mol

1000 cm3 of contained d x 1000 mol

2 x 20/25

= e mol

Conc. of E = e moldm-3

- Molar mass of Na2CO3 = 106 gmol-

Mass of anhydrous Na2CO3 = 106 x X moldm-3

= Wg dm-3

- Mass of water = 13.6 - W

= Z

__Mass of Na2CO3__ = __Molar mass of Na2CO3__

Mass of water (y) Molar mass of water

__W __ = __106__

Z 18y

y = __106 x Z__

W x 18

= m (to the nearest whole number)

- (b)(iii) Molar mass of Na2CO3.yH2O = Concentration in gdm-3 Concentration in moldm-3

= - 13.6

X

= Q gmol- say

Thus (2 x 23) + 12 + (3 x 16) + y(2 + 16) = Q

106 + 18y = Q

y = __Q – 106__

18

= s (to the nearest whole number)