Vapour density of PCI, at 400 K is found to be 90.
Degree of dissociation of PCI, will be
(1) 15.8%
(2) 13.6%
(3) 14.25%
(4) 90%
Answers
answer : option (1) 15.8 %
using formula,
α = (M - m)/M
where α is degree of dissociation, M is theoretical molecular weight and m is observed molecular weight.
theoretical molecular weight of PCl5 = 208.25 g/mol
observed molecular weight = 2 × vapor density
= 2 × 90 = 180 g/mol
now, α = (208.24 - 180)/180 = 28.25/180
= 0.156944
hence, percentage degree of dissociation = α × 100 = 15.69 % it is nearest value of 15.8%
hence, option (1) is correct choice.
also read similar questions : Vapour density of pcl5 is 104.25 at T degree Celsius . Then degree of dissociation of Pcl5 is?
https://brainly.in/question/918771
A 1 litre container contains 2 moles of pcl5 initially if at equilibrium kc is found to be 1 the degree of dissociation ...
https://brainly.in/question/9065437
Given:
- Vapour density of PCl₅ = 90
- Temperature = 400K
To find:
- Degree of dissociation of PCl₅
Solution:
- To calculate the degree of dissociation of PCl₅ in the given specific conditions, we will need to compare it to its theoretical vapor density.
Theoretical vapor density of PCl₅ can be obtained as follows:
- Vapour density = Molecular weight / 2
- We know that molecular weight of PCl₅ = 208.24 g
- Vapour density of PCl = Molecular weight of PCl₅ / 2 = 208.24 / 2 = 104.12
- The vapour density observed at 400K is = 90
Thus,
Degree of dissociation (α) will be = (calculated observation - actual observation) / actual observation
Degree of dissociation (α) = (104.12 - 90)/90 = 0.1569 = 15.69 % Ξ 15.8%
Thus, at 400K with a vapor density of 90, the degree of dissociation of PCl₅ will be (Option 1) 15.8%.