Question

JUST PART B

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 206 torr ; O2, 131 torr ; and He, 110 torr .

Part A What is the total pressure of the mixture? P = 447   torr   SubmitMy AnswersGive Up Correct Significant Figures Feedback: Your answer 446 torrwas either rounded differently or used a different number of significant figures than required for this part. Part B What mass of each gas is present in a 1.00 −L sample of this mixture at 25.0 ∘C? Enter your answers numerically separated by commas.

Answer 1

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 206 torr ; O2, 131 torr ; and He, 110 torr .

total pressure = sum of partial pressure of individual gases =206+131+110 Torr=447 Torr

Mole fraction = moles of gas/total moles of gas

but from gas law, PV=nRT, moles of gas =partial pressure of gas* V/RT

since partial pressure is defined as the pressure exereted if the gas alone occupies the entire volume,

total moles= total pressure of gas*V/RT

hence moles of gas/total moles = mole fraction = partial pressure of gas/total pressure

mole fractions : N2=206/447=0.46, O2= 131/447= 0.29 and He= 110/447= 0.25

total moles of gas can be calculated from n, =PV/RT

P= total pressure in atm= 447/760 atm (760 torr =1atm)=0.59, V= 1L, T= 25deg.c= 25+273= 298K, R= gas constant =0.0821 L.atm/mole.K, n= 0.59*1/(0.0821*298)= 0.024 mole

but mole fraction = moles of gas /total moles

moles of gas =mole fraction* total moles

moles of gases : N2= 0.46*0.024=0.011, O2= 0.29*0.024=0.007 and He= 0.024*0.25=0.006

moles = mass/molar mass, mass =molar mass* moles, molar masses (g/mole): N2= 28, O2=32 and He =40

masses : N2=0.011*28 gm =0.308 gm, O2=0.007*32 =0.224 gm and He=0.006*40 = 0.24 gm