Gaseous N2O4 dissociates into gaseous NO2 according to the reaction N 2 O 4( g ) leftharpoons 2 NO 2( g ) At 300 K and 1 atm pressure, the degree of dissociation of N2O4 is 0.2. If one mole of N2O4 gas is contained in a vessel, then the density of the equilibrium mixture is :

Question

Gaseous N2O4 dissociates into gaseous NO2 according to the reaction N 2 O 4( g ) leftharpoons 2 NO 2( g ) At 300 K and 1 atm pressure, the degree of dissociation of N2O4 is 0.2. If one mole of N2O4 gas is contained in a vessel, then the density of the equilibrium mixture is : (A) 1.56 g/L (B) 3.11 g/L (C) 4.56 g/L (D) 6.22 g/L

Tardigrade Admin · Accepted Answer

P V=n R T ⇒ 1 × V=1 × 0.0821 × 300 ⇒ V=24.63 d=( text mass of mixture / text vol ) =(0.8 × 92+0.4 × 45/24.63)=3.11 gm / lit

Q. Gaseous N2​O4​ dissociates into gaseous NO2​ according to the reaction N2​O4​(g)⇌2NO2​(g) At 300K and 1atm pressure, the degree of dissociation of N2​O4​ is 0.2. If one mole of N2​O4​ gas is contained in a vessel, then the density of the equilibrium mixture is :

1.56 g/L

3.11 g/L

4.56 g/L

6.22 g/L

PV=nRT⇒1×V=1×0.0821×300 ⇒V=24.63 d= vol mass of mixture ​ =24.630.8×92+0.4×45​=3.11gm/ lit

$P V = n RT \Rightarrow 1 \times V = 1 \times 0.0821 \times 300$

$\Rightarrow V = 24.63$

$d = \frac{mass of mixture}{vol}$

$= \frac{0.8 \times 92 + 0.4 \times 45}{24.63} = 3.11 g m /$ lit