Question

A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ∘C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Answer 1

let moles of helium = x

Total moles of mixture = moles of nitrogen +moles of oxygen + moles of helium = 0.25+0.25+x= 0.50+x

Temperature T= 0 deg.c = 0+273.15 K= 273.15K   Pressure, P= 1 atm and Volume, V= 18L

From PV= nRT, R =0.08206 L.atm/mole.K

n= moles of mixture = PV/RT =1*18/ (0.08206*273.15)=0.803 moles

but 0.5+x= 0.803

x= 0.803-0.5= 0.303 moles of helium

Molecular weight of helium = 4 hence mass of helium = moles* Molecular weight = 0.303*4= 1.212 gms