Engineering Entrance Exams Application Date 2025 & Details

Degree of Dissociation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

16 Questions around this concept.

Solve by difficulty

For a concentrated solution of a weak electrolyte (K_eq = equilibrium constant) A₂B₃ of concentration ‘c’, the degree of dissociation ‘ $\alpha$ ’ is

$\left(\frac{K_{e q}}{5 c^4}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{108 c^4}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{25 c^2}\right)^{\frac{1}{5}}$

$\left(\frac{K_{e q}}{6 c^5}\right)^{\frac{1}{5}}$

View Solution

$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons \mathrm{B}_{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g})$ The correct relationship between $\mathrm{Kp}, \alpha$ and equilibrium pressure $\mathrm{P}$ is -

$K p=\frac{\alpha^{\frac{3}{2}} p^{\frac{1}{2}}}{(2+\alpha)^{1 / 2}(1-\alpha)}$

$K_P=\frac{\alpha^{\frac{1}{2}} P^{\frac{3}{2}}}{(2+\alpha)^{3 / 2}}$

$K_P=\frac{\alpha^{\frac{1}{2}} P^{\frac{1}{2}}}{(2+\alpha)^{3 / 2}}$

$K_P=\frac{\alpha^{\frac{1}{2}} P^{\frac{1}{2}}}{(2+\alpha)^{1 / 2}}$

View Solution

Concepts Covered - 2

Degree of Dissociation

Degree of dissociation: It is the extent to which any reactant gets dissociated. It is denoted by $\mathrm{\alpha }$ .

$\mathrm{\alpha=\frac{number \: of\: molecules \: dissociated}{total \: number\: of\: molecules }}$

If "a" be the initial number of moles and the number of moles dissociated be "x" then

$\mathrm{\alpha = \frac{x}{a}}$

Relation between degree of dissociation and Observed Molar mass/Vapor density

Due to dissociation, the total number of moles at equilibrium can be determined. Knowing the number of moles at equilibrium, the observed molar mass can be calculated.

Conversely, if the degree of dissociation is not known but the Observed Vapor density is available, then we can calculate the degree of dissociation.

We know about the law of conservation of mass

$\text{Mass initially taken = Mass after dissociation at equilibrium}$

$\therefore \mathrm{Theoretical\ moles \times Theoretical\ molar\ mass = Observed\ Moles \times Observed\ Molar\ mass}$

Using the above equation, we can calculate the unknown term required to be calculated.

Observed Density and Molar Mass

In equilibrium, the observed molar mass or average molar mass of the reactant is the total mass of the mixture divided by the total number of moles.

$\mathrm{A\: \rightleftharpoons \: nB}$
Initially: 1 0
Equil: 1 - ? n?

$\\\mathrm{M_{observed}\: =\: \frac{Total\: mass\: of\: mix}{Total\: number\: of\: moles\: of\: mix}}\\\\\mathrm{M_{obs}\: =\: \frac{M_{real}}{1-\alpha +n\alpha }\: =\: \frac{M_{real}}{1+\alpha (n-1)}}$

In the equilibrium system, the observed molar mass of the reactant is always different than the actual mass. Thus, when the reaction is reversible, the observed mass varies. In a chemical reaction, some amount of this reactant gets converted into a product, thus observed mass is different from than actual mass.

For example:

$\mathrm{N_{2}O_{_{4}}(g)\: \rightleftharpoons \: \: 2NO_{2}(g)}$
In this reaction, the original molar mass of N₂O₄ = 92g/mol. But the observed molar mass at equilibrium is 80g/mol. The observed molar mass is less than the original molar mass as during the reaction some amount of N₂O₄ is converted into NO₂.

Vapour Density

Similarly, the observed density of the substance is different than the actual density.

Thus, we know:

Vapour density = Molar mass/2

$\\\mathrm{Thus,\: 2\: x\: (V.D)_{obs}\: =\: \frac{2\, x\, (V.D)_{real}}{1+\alpha (n-1)}}\\\\\mathrm{\Rightarrow\: \: d\: =\: \frac{D}{1+\alpha (n-1)}}$