The question was well attempted by candidates and the performance was good In (a), majority of the candidates correctly tabulated the burette readings and used concordant values to calculate the average volume of acid used but some candidates lost marks because
they either altered their burette readings to agree with the supervisor's value or outrightly cancelled their table and retabulated.
In (b)(i), candidates correctly calculated the concentrations of E in moldm-3 and gdm-3 respectively. However, b(ii) and (iii) were difficult to a good number of them while others lost marks because they did not use the required number of significant figures.
The expected solutions to b(i)-(iv) were as follows:
(i) To calculate the concentration of solution E in moldm-3
CDVD = 2
CEVE 1
CE = CDVD
2 x VE
= 0.105 x VD
2 x 20/25
= w moldm-3 (3 sig, fig.)
Alternative Method
1000 cm3 of D contain 0.0105 moles of HCI
:. VD will contain 0.105VD moles of HCI
1000
From the equation for the reaction.
2 moles of HCI == 1 mole of Na2C03
0.105VD moles HCI == 0.105VD moles of Na2C03
1000 2 x 1000
20/25cm3 of E contain 0.105VD moles of Na2C03
2000
1000 cm3 of E will contain 0.1 05VD moles Na2C03
2X20/25
= wmoldm-3
(ii) To calculate the concentration of solution E in gdm-3:
Molar mass of Na2C03 = 46 + 12 + 48 = 106g mol-1
Concentration of Na2C03 = w moldm x 106 gmol-1
= xgdm-3
(iii) |
mass of NaCI in E |
= |
(l0-x) |
|
|
|
= |
Y say |
|
(iv) |
% of NaCI in the mixture |
= |
Y |
x |
100 |
|
|
|
10 |
|
|
|
|
= |
Z% |
|
2NaCI(aq) + C02(g) + H20(l)