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» <html><body><p>`1.4ppm`<br/>`14ppm`<br/>`2.8ppm`<br/>`6.25ppm`</p>Solution :`SO_(2)+H_(2)O <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> H_(2)SO+_(3)` <br/> `underset(+7)(2MnO_(4)^(-))+underset(+4)(2H_(2)SO_(3)) rarr underset(+6)(5H_(2)SO_(4))+underset(+2)(2Mn^(2+))` <br/> `2 mol <a href="https://interviewquestions.tuteehub.com/tag/es-446539" style="font-weight:bold;" target="_blank" title="Click to know more about ES">ES</a> MnO_(4)^(-) = 5 mol e H_(2)SO_(3) or SO_(2)` <br/> `2.5xx10^(-5)` mole `MnO_(4)^(-) = 5/2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2.5xx10^(-5) mol es SO_(2)` <br/> `=6.25xx10^(-5)mol es SO_(2)` <br/> At STP, `6.25xx10^(-5) mol es = 6.25xx10^(-5)xx22.4L SO_(2)` <br/> `=1.4xx10^(-3)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>` <br/> In `1000L` air, `SO_(2)=1.4xx10^(-3)L` <br/> In `10^(6)L(ppm), SO_(2) = (1.4xx10^(-3)xx10^(6))/(10^3) = 1.4ppm`.</body></html>