Question

Vapour density is found to be 100 when 1 mole of pcl5 is taken in 10 l flast at 27c the % degree of dissociation of pcl5 is

Answer 1

Answer:

4.25%.

Explanation:

We get the reaction of decomposition of PCl5 as PCl5 -> PCl3 + Cl2. if we take the % of dissociation to be a. Hence, we know that a=Mf - Mi/Mf(n-1) where Mf and Mi are the molar mass before and after dissociation and n is the number of moles. So, Mi = 31 + 5*35.5 = 208.5 and Mf = 2 * vapour density which is 2*100=200 and the n=2.

So, after substitution we will get that 208.5-200/208.5(2-1) which on solving we will get 0.0425. So, the dissociation % will be 0.0425*100 which will be 4.25%.

Answer 2

Answer:

We get the reaction of decomposition of PCl5 as PCl5 -> PCl3 + Cl2. if we take the % of dissociation to be a. Hence, we know that a=Mf - Mi/Mf(n-1) where Mf and Mi are the molar mass before and after dissociation and n is the number of moles. So, Mi = 31 + 5*35.5 = 208.5 and Mf = 2 * vapour density which is 2*100=200 and the n=2.

So, after substitution we will get that 208.5-200/208.5(2-1) which on solving we will get 0.0425. So, the dissociation % will be 0.0425*100 which will be 4.25%.

Explanation: