Question

so3 is heated in a closed vessel an equilibrium 2so3 = 2so2 + o2 is established the vapour density of the mixture in which so3 is 50 % dissociated is

Answer 1

2so3=2so2+o2

We have to find vapour density...so

D/d=1+(n-1)alpha

MM of so3 is 80

D=MM/2=40

ALPHA=50/100

n=3/2

PUTTING THE VALUES IN FORMULA WE GET...

40/d=5/4

d=32..ans

Answer 2

Given:

$2SO_3$⇔$2SO_2+O_2$

To find:

Vapor density = ?

Formula to be used:

$\alpha = \frac{D-d}{(n-1)d}$

Calculation:

Vapor density of 100% dissociation (D) = $\frac{80}{2} = 40$

From the given equation it is inferred that 2 molecules of $SO_3$ on dissociation gives 3 molecules of gaseous products.

∴ 1 molecule of $SO_3$ on dissociation gives = $\frac{3}{2} = 1.5 = n$

Substituting the obtained value in the following formula, gives vapor density

$\alpha = \frac{D-d}{(n-1)d}$

$0.5 = \frac{40-d}{(1.5-1)d}$

$0.5 = \frac{40-d}{0.5d}$

0.25 d = 40 - d

1.25d = 40

d = 32

Conclusion:

The vapor density of the mixture in which $SO_3$ is 50 % dissociated is 32.

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