Question

$1000\text{L}$ of air at STP was dissolved in water and it required $2.5\times {10}^{-5}$ moles of $KMn{O}_4$ for complete reaction with the pollutant $S{O}_2$. Thus, the $S{O}_2$ content in air is:

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

Answer 1

The correct option is A 1.4 ppm

$S{O}_2+H_{2}O\to H_{2}S{O}_3$
$2Mn{O}_4^{-}+5H_{2}S{O}_3\to 2M{n}^{2+}+3H_{2}O+5S{O}_4^{2-}+4H^{+}$
$2molesMn{O}_4^{-}=5molesof{H}_{2}S{O}_{3}orS{O}_2$
$2.5\times {10}^{-5}molesMn{O}_4^{-}=\frac{5}{2}\times 2.5\times {10}^{-5}molesS{O}_2$
$=6.25\times {10}^{-5}molesS{O}_2$
At S.T.P, $6.25\times {10}^{-5}moles=6.25\times {10}^{-5}\times 22.4LS{O}_2$
$=1.4\times {10}^{-3}L$
In 1000 L air, $S{O}_2=1.4\times {10}^{-3}L$
$In{10}^{6}L,S{O}_2=\frac{1.4\times {10}^{-3}\times {10}^6}{{10}^{-3}}=1.4ppm$