What is the vapour density of mixture of `PCL_(5)` at `250^(@)C` when it has dissociated to the extent of `80%` ?

An unknown compound A dissociates at 500^(@)C to give products as follows - A ( g) hArr 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-

An unknown compound A dissociates at 500^(@)C to give products as follows - A ( g) hArr 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-

The Vapour Density of mixture of PCl_(5), PCl_(3) and Cl_(2) is 92. Find the degree of dissociation of PCl_(5) .

At temperature T K PCl_5 is 50% dissociated at an equilibrium pressure of 4 atm. At what pressure it would dissociate to the extent of 80% at the same temperature ?

At temperature T k PCI_5 is 50% dissociated at an equilibrium pressure of 4 atm. At what pressure it would dissociate to the extent of 80% at the same temperature ?

When PCl_(5) is heated, it dissociates into PCl_(3) and Cl_(2) . The vapour density of the gas mixture at 200^(@)C and at 250^(@)C is 70 and 58 , respectively. Find the degree dissociation at two temperatures.

When PCl_(5) is heated, it dissociates into PCl_(3) and Cl_(2) . The vapour density of the gas mixture at 200^(@)C and at 250^(@)C is 70 and 58 , respectively. Find the degree dissociation at two temperatures.

At 250^(@)C and 1 atmospheric pressure, the vapour density of PCl_(5) is 57.9 . What will be the dissociation of PCl_(5) –

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5) at 250^(@)C is numerically three times of K_(p) .