Question

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.)

Answer 1

Here mass of the substance taken `=0.30gm`
Atmospheric pressure `=715 mm` Hg
Vapour pressure of water at `300`
`K=15mm`
Volume of notrogen collected `=50ml`
Room temperature `=300K`
Actual pressure of the gas (dry gas) `=715-15=700mmHg`
To convert the volume at experimental conditions to volume at STP.
`{:("experiment at 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`
`=(28xx41.9xx100)/(22400xx0.3)`
`=17.46`