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Consider a room that is 20 xx 10 m with 15 m eciling. If pollutant present is 2 ppm, how many grams of pollutant are present in this room. (Density of air = 1 gm//"litre"

Answer»


Solution :`V = L xx B xx H = 20 xx 10 xx 15`
`= 3000 m^(3) = 3000 xx 10^(3)` litre
`= 3 xx 10^(6)` litre
2 grams of pollutant in `10^(6)` grams of air
x grams of pollutant in `3 xx 10^(6)` GRAM of air`""` (because `d = 1 gm//"litre"` )
`x = (3XX 10^(6) xx 2)/(10^(6)) = 6`


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