Question
Question: 12 g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of sodium bicarb...
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 by 250 ml. 25 ml of this solution was completely oxidized 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 (Na2S2O3.5H2O) in one litre. Calculate the percentage of arsenious oxide in the sample. (Atomic mass of As = 75)
Solution
Normality is generally used to measure the concentration of a solution. It is represented by the letter N and sometimes referred to as the equivalent concentration of a solution. Units of normality expressed as eqL−1 or meqL−1.
Complete Solution :
Normality can also be defined as the number of gram or mole equivalents of solute present in one litre of a solution. Where equivalent represents the number of moles of reactive units in a compound.
Normality is calculated by the formula: N=equivalent weight×volumeWeight of solute
Normality of Na2S2O3.5H2O solution = 248×124.8=0.1N
Applying the normality equation:
N1V1=N2V2
Volume of AS2O3solution in NaHCO3×Normality = Volume of iodine solution×Normality
- We have to find out normality of AS2O3solutioninNaHCO3, let us suppose it as N
25×N=22.4×0.1
N=2522.4×0.1=0.0896
Amount of As2O3 present in 250 ml of solution is given by:
N×1000Equivalent mass of As2O3×250
0.0896×4×1000198×250=1.1088g
Percentage of As2O3 = 121.1088×100=9.24
Percentage of arsenious oxide in the sample is 9.24%.
Note: Arsenious oxide is an inorganic compound. It is generally used as a medication to treat a type of cancer called acute promyelocytic leukemia in which it is by injection directly to the vein. It is also used in the manufacture of wood preservatives, pesticides etc.