Question

An unknown compound a dissociates at 500 c to give products as follows a (g) givesb(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

Answer 1

Answer:

The answer is 120 g/mol

Explanation:

The relation of vapour density with dissocation constant is:

α=(D−d)/(n−1)d

α = degree of dissociation = 10 % = 0.10

D = Initial vapor density  

d = vapor density at equilibirum = 50

n = moles of gaseous product = 3

So, α=(D−d)/(n−1)d

0.10 =D−50/(3−1)50

0.10 = D−50/100

10 =D-50

10 = D-50

D = 60

Vapor density = 60

Molar mass = 2x vapor density = 2 x 60 = 120 g/mol

Molar mass of A = 120 g/mol

Answer 2

Answer:

120g/mol

Explanation:

it is the best for me to come over