A sample of air analyzed at 0{ }^{\circ} {C} and 1 atm pressure is reported to contain 0.02 {ppm} (parts per million) of {NO}_{2}. Assume the gram molecular mass of {NO}_{2} as 46 and its volume at 0^{\circ} {C} and 1 atm pressure as 22.4 litres per mole. The equivalent {NO}_{2} concentration (in microgram per cubic meter, round off to 2 decimal places) would be __________

GATE Civil Engineering (CE) 2019 Shift 1 Solved Paper

Question : 40 of 65

Marks: +1, -0

A sample of air analyzed at

0{ }^{∘} {C}

and 1 atm pressure is reported to contain

0.02 {ppm}

(parts per million) of

{NO}_{2}

. Assume the gram molecular mass of

{NO}_{2}

as 46 and its volume at

0^{∘} {C}

and 1 atm pressure as

22.4

litres per mole. The equivalent

{NO}_{2}

concentration (in microgram per cubic meter, round off to 2 decimal places) would be __________

Solution:

1{ppm} {NO}_{2}=\;{1 \;\text " part of "\; {NO}_{2}}/{10^{6} \;\text " parts of air "\;}=\;{1 {m}^{3} \;\text " of "\; {NO}_{2}}/{10^{6} {m} \;\text " of air "\;}

1 mole of

{NO}_{2}

occupies

22.4

litres of volume at STP.
⇒1 mole of

{NO}_{2}

occupies

0.0224 {m}^{3}

volume at STP.

⇒ 1 {m}^{3}

{NO}_{2}

contains

\;{1}/{0.0224}

moles

=44.64

moles
1 mole of

{NO}_{2}=46 {gms}

⇒ 1 {m}^{3}

{NO}_{2}=44.64 × 46 {gms}=2053.57 {gms}

⇒ 0.02 {m}^{3}

{NO}_{2}=41.07 {gms}

{NO}_{2}

∴ \;\; 0.02 {ppm}

{NO}_{2}=\;{0.02 {m}^{3} \;\text " of "\; {NO}_{2}}/{10^{6} {m}^{3} \;\text " of air "\;}=41.07 {gms}

per

10^{6} {m}^{3}

=41.07 × 10^{-6} {gm} {/} {m}^{3}=41.07 µ {g}{/} {m}^{=}

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