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Question

12 g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of sodium bicarbonate and the resulting solution was diluted to 250 mL. 25 mL of this solution was completely oxidised by 22.4 mL of a solution of iodine. 25 mL of this iodine solution reacted with same volume of a solution containing 24.8 g of sodium thiosulphate solution (Na2S2O35H2O) in one litre. Calculate the percentage of arsenious oxide in the sample. (Atomic mass of As=75).

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Solution

Normality of Na2S2O3soln.=24.8248=0.1 N
Applying N1V1=N2V2
Volume of As2O3 soln. in NaHCO3× Normality of this soln.
= Volume of iodine soln. × Normality of iodine soln.
25×N1=22.4×0.1
N1=22.4×0.125
Amt. of As2O3 present in 250 mL of the solution
=N1=Equivalent mass of As2O31000×250
=22.4×0.125×1984×2501000=1.1088 g
Percentage of As2O3=1.108812×100=9.24.

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