Question

The degree of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction,
$P{ Cl }_{ 5 }\quad \rightleftharpoons \quad P{ Cl }_{ 3 }\quad +\quad { Cl }_{ 2 }$
Assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at 400 K and 1.0 atmosphere. (Relative atomic mass of P = 31.0 and Cl = 35.5)
a.) 4.60 g/litre
b.) 4.53 g/litre
c.) 4.75 litre
d.) 4.35 litre

Answer 1

Hint: Degree of dissociation is the fraction of the solute that has been dissociated. The density of a mixture is defined as the mass of the mixture per unit volume of the mixture.

Complete step by step solution:
The degree of dissociation can be defined as the amount of solute that is dissociated into ions or radicals per mole.

Now, considering the question given.
$P{ Cl }_{ 5 }\quad \rightleftharpoons \quad P{ Cl }_{ 3 }\quad +\quad { Cl }_{ 2 }$
Let us assume that initially there was 1 mole of $P{Cl}_{5}$. After the reaction, it is given that the degree of dissociation is 0.4 at temperature = 400 K and pressure = 1 atm. This means that after the reaction the moles of $P{Cl}_{5}$ left = $1-0.4$ moles = $0.6$ moles. And the moles of products $P{Cl}_{3}$ and ${Cl}_{2}$ are $0.4$ each.
Therefore, the total no. of moles = $0.6 + 0.4 + 0.4 = 1.4$.
Now, according to Ideal gas equation,
$PV\quad =\quad nRT$
where, P = pressure, V=volume, n=total no. of moles, R=universal gas constant= $0.082\quad L-atm/K-mol$ and T=temperature
We can find the volume using the above equation as,
$V\quad =\quad \cfrac { nRT }{ P }$
Now, n=1.4, T=400K, P=1atm and R= $0.082\quad L-atm/K-mol$. Substituting the values in above equation, we get,
$V\quad =\quad \cfrac { 1.4\quad \times \quad 0.082\quad \times \quad 400 }{ 1 }$
$\implies V\quad =\quad 45.976\quad L$
Now, the mass of $P{Cl}_{5}$ = 31 + 5x35.5 = 208.5g/mole. Thus for 1 mole, the mass = 208.5g.
The density is given as,
$Density\quad =\quad \cfrac { Mass }{ Volume }$
Substituting the value of mass and volume, we get
$Density\quad =\quad \cfrac { 208.5 }{ 45.976 }$
$\implies Density = 4.53 g/litres$
Hence, option (b) is the correct option.

Note: R is a universal gas constant, It has various values in different units. And we can use those values according to our convenience. But in this question we required that value of R which was $0.082\quad L-atm/K-mol$.