Question

At ${200}^o$C a compound 'A' (molar mass = 100g $mo{l}^{-1}$) in the vapour phase dissociates as
A(g) $\to$ 2B (g) + C(g).
The vapour density of the vapour at ${200}^o$C is found to be 30. The percent dissociation of the compound A at ${200}^o$C will be:

• A. 33 %
• B. 66 %
• C. 87 %
• D. 100 %

Answer 1

The correct option is A 33 %

Given,

$A\left(g\right)⟶2B\left(g\right)+C\left(g\right)$

No. of moles of gaseous product, according to reaction

$=2+1=3⟶\left(2\right)$

Also, Molar mass of $A=100gmo{l}^{-1}$

$\Rightarrow$ V.O=Vapour density of $A$ initially=$\frac{100}{2}=50⟶\left(3\right)$ [$\because Molarmass=VO\times 2$]

At equilibrium,

it is given Vapour density becomes $30⟶\left(4\right)$

We have to find degree of dissociation of $A$ i.e; $\alpha$

We know,

$\alpha =\frac{(O-d)}{(n-1)d}⟶\left(1\right)$ is the relation between degree of dissociation and vapour densities

where, $\alpha =$ degree of dissociation, $d=$ vapour density at equilibrium

$O=$initial vapour density, $n=$ moles of gaseous product

From, (1),(2),(3),(4) $\alpha =\frac{50-30}{(3-1)30}=\frac{20}{2\times 30}=\frac{1}{3}=0.333$

$\therefore$ If degree of dissociation $\alpha =0.333$

$\Rightarrow$ Percentage of dissociation=$\alpha \times 100$

$=33.33$%