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Degree of Dissociation MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

8 Questions around this concept.

Solve by difficulty

EASY

If the equilibrium constant for

$N_2(g) + O_2(g)\rightleftharpoons 2NO(g)$ is K, the equilibrium constant for $1/2N_2(g) + 1/2O_2(g)\rightleftharpoons NO(g)$ will be

$K^{1/2}$

$\frac{K}{2}$

$K$

$K^2$

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200Ml of $\mathrm{O}_2(\mathrm{~g})$ effuse from a Porous container in 200 sec .75 Ml of unknowns gas effuse under the same condition of temperature and pressure in 225sec. Calculates vapour density of unknown gas?

128

144

View Solution

Concepts Covered - 2

Degree of Dissociation

Degree of dissociation: It is the extent to which an electrolyte gets dissociated in a solvent. It is shown by 𝛂.

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

Degree of dissociation(𝛂) depends on the following factors:

Nature of solute and solvent: For strong electrolytes, 𝛂 is more than that for weak electrolytes.
𝛂 ∝ Dielectric constant of solvent
That is, greater the dielectric constant of a solvent more will be ionization of electrolyte in it.
The degree of dissociation of weak electrolyte ∝ Dilution that is 𝛂 is maximum at infinite dilution.
𝛂 ∝ 1/Concentration
𝛂 ∝ Temperature

Observed Density and Molar Mass

In equilibrium, 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 - 𝛂 n𝛂

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

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

For example:

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

Vapour Density

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

Thus, we know:

Vapour density = Molar mass/2

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