1000L of air a STP was dissolved in water and required 2.5xx10^(-5) moles of KMnO_(4) for complete reaction of SO_(2) as pollutants. Thus , SO_(2) content in air is

1000L of air a STP was dissolved in water and required 2.5xx10^(-5) mo

1.	1000L of air a STP was dissolved in water and required 2.5xx10^(-5) moles of KMnO_(4) for complete reaction of SO_(2) as pollutants. Thus , SO_(2) content in air is
Answer» `1.4ppm` `14ppm` `2.8ppm` `6.25ppm` Solution :`SO_(2)+H_(2)O RARR H_(2)SO+_(3)` `underset(+7)(2MnO_(4)^(-))+underset(+4)(2H_(2)SO_(3)) rarr underset(+6)(5H_(2)SO_(4))+underset(+2)(2Mn^(2+))` `2 mol ES MnO_(4)^(-) = 5 mol e H_(2)SO_(3) or SO_(2)` `2.5xx10^(-5)` mole `MnO_(4)^(-) = 5/2 XX 2.5xx10^(-5) mol es SO_(2)` `=6.25xx10^(-5)mol es SO_(2)` At STP, `6.25xx10^(-5) mol es = 6.25xx10^(-5)xx22.4L SO_(2)` `=1.4xx10^(-3)L` In `1000L` air, `SO_(2)=1.4xx10^(-3)L` In `10^(6)L(ppm), SO_(2) = (1.4xx10^(-3)xx10^(6))/(10^3) = 1.4ppm`.