`0.30` gm of an organic compound gave `50` mL of nitrogen collected at `300` K and `715` mm pressure in dumas method. Calculate the percentage of nitrogen in the compound (Vapour pressure of water or aqueous tension of water at `300` K is `15` mm).

Explanation

Here mass of the substance taken `=0.30gm`
Atmospheric pressure `=715` mm Hg
Vapour pressure of water at `300` K `=15` mm
Volume of nitrogen collected `=50` mL
Room temperature `=300` K
Actual pressure of the gas (dry gas) `=715-15=700` mm Hg
To convert the volume at experimental conditions to volume at STP.
`{:("experimental condition, At STP"),(P_1=700mm,P_2=760mm),(V_1=50ml,V_2=?),(T_1=300K,T_2=273K):}`
Substituting these values in the gas equation
`=(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`, we get
`(700mmxx50ml)/(300K)=(760mmxxV_2ml)/(273K)`
or `V_2=(273xx700xx50)/(300xx760)=41.9ml`
Step 2 : Percentage of nitrogen `=(28)/(22400)xx("Vol. of "N_2" at STP")/("Mass of compound")xx100`
Percentage of nitrogen `=(28xx41.9xx100)/(22400xx0.3) =17.46`