2.3 grams of a mixture of NO2 and N2O4 have a pressure of 0.82 atm at temperature, T K and volume, V litres. If VT=1300magnitude, calculate PNO2. Assume that all the NO2 was formed N2O4.

(A) 0.52 atm                             

(B) 0.38 atm

(C) 0.19 atm                             

(D) 0.41 atm

Concept Videos :-

#5 | Dalton's Law of Partial Pressure

Concept Questions :-

Dalton's Law

D.    Using PV=mM RT, PVT=mM×R,           0.82×1300=2.4M×0.0821           2.3×0.08210.8230069            If one mole of N2O4 yields by dissociation,             1  x mole of N2O4 and 2x mole of NO2,              The total number of moles = 1 + x with a mean molecular weight of 69.             1+x69=1×92              x=92-6969=2369=13              PNO2=2x1+x×0.82             =2×1343 × 0.82 = 0.41 atm

Difficulty Level:

  • 20%
  • 38%
  • 12%
  • 32%