NTA Abhyas NEET2020ChemistryChemical EquilibriumPractice
An unknown compound A dissociates at 500 ° C to give products as follows – A ( g ) ⇌ B ( g ) + C ( g ) + D ( g ) Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10 % . What will be the molecular weight of Compound A–
Options
- A120
- B130
- C134
- D140
Correct answer
A. 120
Step-by-step solution
A g ⇌ B g + C g + D g α = Iintial vapour density − Vapour density at equilibrium moles of gaseous product × vapour density at equilbrium 1 10 = Vapour density − 50 3 − 1 50 Vapour density = 60 Molar mass of compound A = 2 × VD = 2 × 60 = 120 gm / mol