(1a)
DRAW A TABLE WITH THE FOLLOWING

Burette reading|ROUGH|1st |2nd |3rd
Final |13.60|15.50|14.80|12.50
Initial |1.00 |3.10 |2.40 |0.00
Vol. acid used |12.60|12.60|12.40|12.50

Volume of S2O3²- =12.60+12.40+12.50/3
=37.50/3
=12.50cm³

(1bi)
Conc. in g/dm³ = molar mass * conc in mol/dm³
But molar mass of Na2S2O3
=(23*2)+(32*2)+(16*3)
=46 + 64 + 48
=158g/mol
Therefore 15.8g/dm³ = 158g/mol * conc of A in mol/dm³

Conc of A in mol/dm³ = 15.8/158
=0.1mol/dm³

(1bii)
Using CaVa/CbVb = na/nb
0.1*12.50/Cb * 25.00 = 2/1

50Cb = 1.25
Cb = 1.25/50
Cb = 0.025mol/dm³

The conc of I2 in B = 0.025mol/dm³

(1biii)
Gram conc of I2 = molar mass * molar conc
= (127*2)*0.025
=6.35g/dm³

Percentage mass of I2 in sample = 6.35/9.0 *100%
= 0.7056 * 100%
=70.56%

(1c)
There have to be a change in the colour of the mixture before it is added. The end point is known when the blue colour formed as starch is added, changes to colourless
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(2a)
Test: C + Water and filtered

Observation: part of C dissolves, colour less filtrate is formed

Inference: C is a mixture of a soluble salt

(2bi)
Test: 2cm^2 of filterate of salt C + AgNO3(aq) then dilute HNO3

Observation: white precipitate is formed

Inference: Cl^- present

(2bii)
Test: Solution im 2(b)(i) + excess NH3 solution

Observation: precipitate disorder in excess solution of NH3

Inference; Cl^- confirmed

(2ci)
Test: Resident + dilute HCl + shake

Observation; residue dissolved to form a blue solution

Inference; Cu^2+ present

(2cii)
Test; solution in 2(c)(i) + NH3 solution in drops and then in excess

Observation: light pale blue precipitate is formed precipitate dissolved in excess solution of NH3 to give a deep blue solution