Question

Given \[COC{l_2}\]​ gas dissociates according to the equation, \[COC{l_2}\left( g \right) \rightleftharpoons CO\left( g \right) + C{l_2}\left( g \right)\] When heated to \[700\;K\;\] the density of the gas mixture at \[1.16\;atm\;\] and at equilibrium is \[1.16\;g/litre\]. The degree of dissociation of \[C{O_2}\] at \[700\;K\;\] is
A.$0.28$
B.$0.50$
C.$0.72$
D.$0.42$

Answer 1

Hint: We need to understand the concept of degree of dissociation along with its formula and calculate it using the given values. In chemistry and biochemistry, dissociation is a reversible mechanism in which molecules (or ionic compounds such as salts or complexes) detach or break into other things such as atoms, ions, or radicals. As an acid dissolves in water, heterolytic fission breaks a covalent bond between an electronegative atom and a hydrogen atom, resulting in a proton and a negative ion. The opposite of association or recombination is dissociation.

Complete answer:
We must have to know that the fraction of a substance's molecules dissociating at a given time is known as its degree of dissociation. The number of moles of \[N{H_3}\] dissociated per mole of \[N{H_3}\] is known as the degree of dissociation (\[a\]) of \[N{H_3}\]. The degree of dissociation of \[N{H_3}\] is \[\dfrac{x}{a}\] if \[x\] moles of \[N{H_3}\] dissociate from 'a' moles of \[N{H_3}\].
It is denoted as a= \[\dfrac{{{M_T} - {M_O}}}{{(n - 1){M_O}}}\]where MT is the theoretical mass value and MO is the observed mass value which is calculated as ${M_O} = \dfrac{{dRT}}{P}$
For the given question, ${M_O} = 57.47$g with the given values of density, temperature, pressure and the universal gas constant. MT For $COC{l_2}$ is calculated (using the atomic masses) as $99g$. Putting these values in the degree of dissociation formula,
a= \[\dfrac{{{M_T} - {M_O}}}{{(n - 1){M_O}}}\] where n is the number of moles of products in gaseous state (=2)
   \[ = \dfrac{{99 - 57.47}}{{57.47}} = 0.72\]

Therefore, the correct option is option “c”.

Note:
We must note that the fraction of original solute molecules that have dissociated is the dissociation degree displaystyle alpha. The Greek symbol is commonly used to represent it. The amount of solute dissociated into ions or radicals per mole is referred to as degree of dissociation. The degree of dissociation would be equivalent to 1 in the case of very strong acids and bases. The degree of dissociation would be lower for weaker acids and bases. This parameter and the van 't Hoff factor I have a straightforward relationship. If a solute material dissociates into n ions, it is said to have dissociated into n ions, then \[i = 1 + \alpha \left( {n - 1} \right).\]