(a)

	Rough Tree	1st Titre	2nd Titre	3rd Titre
Final reading cm\(^3\)	12.60	25.10	12.40	25.00
Initial reading cm\(^3\)	0.60	12.60	0.00	12.40
Volume of acid cm\(^3\)	12.60	12.50	12.40	12.60

Average Titre = \(\frac{12.50 + 12.40 + 12.60}{3}\) = 12.50cm\(^3\)
Equation of the reaction I\(_2\) + 2S\(_2\)O\(_3^{2-}\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\)

(b)(i) Concentration of A in moldm\(^{-3}\)
= \(\frac{\text{Concentration of A in gdm}^{-3}}{\text{Molar mass of A in g mol}^{-1}}\)
= \(\frac{15.8gdm^{-3}}{\text{Molar mass of A}}\)
Molar mass Na\(_2\)S\(_2\)O\(_3\) = (23)2 + (32) 2 + (16)3
= 46 + 64 + 48
= 158 g mol\(^{-1}\)
concentration of A in moldm\(^{-3}\)
= \(\frac{15.8gdm^{-3}}{158gmol^{-1}}\) = 0.100 moldm\(^{-3}\)

(ii) Concentration of I\(_2\) in B moldm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
Where;
C\(_A\) = 0.100 moldm\(^{-3}\)
V\(_A\) = 12.50 cm\(^2\)
C\(_B\) = ?
V\(_B\) = 25.00 cm\(^2\)
C\(_B\) = \(\frac{0.100 \times 12.50 \times 1}{2 \times 25.00}\)
C\(_B\) = \(\frac{1.25}{50}\)
C\(_B\)= 0.0250 moldm\(^{-3}\)

(iii) percentage by ass of I\(_2\) in the sample
= \(\frac{\text{Concentration in gdm}^{-3}}{9.0 gdm^{-3} \text{of impure}} \times \frac{100%}{1}\)
Concentration in gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{ Molar mass of } I_2}\)
0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{Molar mass of } I_2}\)
I\(_2\) = 127 x 2 = 254/mol
0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} (Pure)}{ \text{Molar mass of }I_2}\)
Concentration in gdm\(^{-3}\) (Pure) = 0.0250 x 254
= 6.35 gdm\(^{-3}\)
Percentage by mass of I\(_2\) in the sample
= \(\frac{6.35}{9.0} \times \frac{100%}{1}\) = 70.555
= 70.56%

(c) Starch indicator was not added to the titration mixture at the beginning of the titration in order to obtain an accurate end-point or to prevent the formation of the complex which reduces the accuracy of the titre value.