`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

Options

A. `1.4ppm`
B. `14ppm`
C. `2.8ppm`
D. `6.25ppm`

Answer

A

Explanation

`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`.