Nitrogen dioxide ( NO 2) cannot be obtained in a pure form in the gas phase at exists as a mixture of NO 2 and N 2 O 4. At 298 K and 0.98 atm, the density of this gas mixture is 2.764 g L -1. Thus partial pressure of NO 2 in the mixture is

Question

Nitrogen dioxide ( NO 2) cannot be obtained in a pure form in the gas phase at exists as a mixture of NO 2 and N 2 O 4. At 298 K and 0.98 atm, the density of this gas mixture is 2.764 g L -1. Thus partial pressure of NO 2 in the mixture is (A) 0.40 atm (B) 0.51 atm (C) 0.49 atm (D) 0.60 atm

Tardigrade Admin · Accepted Answer

pV = nRT =(w/m) R T ∴ p=(w/V) m R T p=(d/m) R T m (mixture ) = (d R T/p) =(2.764 × 0.0821 × 298/0.98) =69 g mol -1 Let mole fraction of NO 2=x and that of N 2 O 4=(1-x) Then (M( NO 2) x+M( N 2 O 4)(1-x)/1+(1-x)) 46 x+92(1-x)=69.0=69.0 g mol -1 Thus, x=0.50 p NO 2=0.50 × p text total =0.49 atm

Q. Nitrogen dioxide (NO2​) cannot be obtained in a pure form in the gas phase at exists as a mixture of NO2​ and N2​O4​. At 298K and 0.98atm, the density of this gas mixture is 2.764gL−1. Thus partial pressure of NO2​ in the mixture is

0.40 atm

0.51 atm

0.49 atm

0.60 atm

pV=nRT=mw​RT ∴p=Vw​mRT p=md​RT m (mixture ) =pdRT​ =0.982.764×0.0821×298​ =69gmol−1 Let mole fraction of NO2​=x and that of N2​O4​=(1−x) Then 1+(1−x)M(NO2​)x+M(N2​O4​)(1−x)​ 46x+92(1−x)=69.0=69.0gmol−1 Thus, x=0.50 pNO2​​=0.50×ptotal ​=0.49atm

Q. Nitrogen dioxide $(N O_{2})$ cannot be obtained in a pure form in the gas phase at exists as a mixture of $N O_{2}$ and $N_{2} O_{4}$ . At $298 K$ and $0.98 a t m$ , the density of this gas mixture is $2.764 g L^{- 1}$ . Thus partial pressure of $N O_{2}$ in the mixture is