Answer :
Final answer:
The degree of dissociation of PCl5 when the vapour density drops from 104.25 to 62 can be calculated by equating the average molar mass after dissociation to twice the observed vapour density. This molar mass is determined considering the degree of dissociation 'n' and the molar masses of PCl5, PCl3 and Cl2.
Explanation:
The vapour density of a gas is directly proportional to its molar mass. In the case of PCl5, when it is heated, it dissociates into PCl3 and Cl2. The initial vapour density of PCl5 is given as 104.25. However, when it dissociates upon heating, the observed vapour density reduces to 62. This reduction in vapour density is due to the reduction in molar mass of the substance because of dissociation. To calculate the degree of dissociation, we assume it to be 'n'. According to the equation PCl5 ⇌ PCl3 + Cl2, one mole of PCl5 dissociates to give one mole each of PCl3 and Cl2. So, the total number of moles after dissociation is 1 - n + n + n = 1 + n. Therefore, the average molar mass after dissociation is calculated by = [(1-n)M1 + nM2 + nM3] / (1 + n), where M1, M2 and M3 are the molar masses of PCl5, PCl3 and Cl2, respectively.
Substituting all values and equating the calculated average molar mass to twice the observed vapour density (since vapour density is half the molar mass), we can determine the value of 'n' which represents the degree of dissociation.
Learn more about Vapour Density and Degree of Dissociation here:
https://brainly.com/question/35913690
#SPJ11
