JEE Main 2023 — Chemistry Physical Chemistry

2023mcqmedium

For a concentrated solution of a weak electrolyte ( ${K}_{\mathrm{eq}}=$ equilibrium constant) ${A}_{2}{B}_{3}$ of concentration $‘C’,$ the degree of dissociation $‘\alpha ’$ is

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Options

A
${(\frac{{K}_{\mathrm{eq}}}{5{c}^{4}})}^{\frac{1}{5}}$
B
${(\frac{{K}_{\mathrm{eq}}}{108{c}^{4}})}^{\frac{1}{5}}$
C
${(\frac{{K}_{\mathrm{eq}}}{25{c}^{2}})}^{\frac{1}{5}}$
D
${(\frac{{K}_{\mathrm{eq}}}{6{c}^{5}})}^{\frac{1}{5}}$

Did you get this right?

Solution

Given that concentration of ${A}_{2}{B}_{3}$ is $C$ and degree of dissociation is $\alpha$

The equilibrium reaction of weak electrolyte ${A}_{2}{B}_{3}$ is

$\begin{matrix}{A}_{2}{B}_{3} & \rightleftharpoons 2{A}^{3+} & +3{B}^{2-} \\ C & - & - \\ \downarrow & \downarrow & \downarrow \\ C(1-\alpha ) & 2C\alpha & 3C\alpha \end{matrix}$

Now, The equilibrium constant is the ratio of the concentrations of reactants and products raised to the stoichiometric coefficients.

${K}_{\mathrm{eq}}=\frac{(2C\alpha {)}^{2}(3C\alpha {)}^{3}}{C(1-\alpha )}=108{C}^{4}{\alpha }^{5}$

$\frac{{K}_{\mathrm{eq}}}{108{C}^{4}}={\alpha }^{5}\Rightarrow \alpha ={(\frac{{K}_{\mathrm{eq}}}{108{C}^{4}})}^{\frac{1}{5}}$

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JEE Main 2023 — Chemistry Physical Chemistry

2023mcqmedium

For a concentrated solution of a weak electrolyte ( ${K}_{\mathrm{eq}}=$ equilibrium constant) ${A}_{2}{B}_{3}$ of concentration $‘C’,$ the degree of dissociation $‘\alpha ’$ is

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Options

A
${(\frac{{K}_{\mathrm{eq}}}{5{c}^{4}})}^{\frac{1}{5}}$
B
${(\frac{{K}_{\mathrm{eq}}}{108{c}^{4}})}^{\frac{1}{5}}$
C
${(\frac{{K}_{\mathrm{eq}}}{25{c}^{2}})}^{\frac{1}{5}}$
D
${(\frac{{K}_{\mathrm{eq}}}{6{c}^{5}})}^{\frac{1}{5}}$

Did you get this right?

Solution

Given that concentration of ${A}_{2}{B}_{3}$ is $C$ and degree of dissociation is $\alpha$

The equilibrium reaction of weak electrolyte ${A}_{2}{B}_{3}$ is

$\begin{matrix}{A}_{2}{B}_{3} & \rightleftharpoons 2{A}^{3+} & +3{B}^{2-} \\ C & - & - \\ \downarrow & \downarrow & \downarrow \\ C(1-\alpha ) & 2C\alpha & 3C\alpha \end{matrix}$

Now, The equilibrium constant is the ratio of the concentrations of reactants and products raised to the stoichiometric coefficients.

${K}_{\mathrm{eq}}=\frac{(2C\alpha {)}^{2}(3C\alpha {)}^{3}}{C(1-\alpha )}=108{C}^{4}{\alpha }^{5}$

$\frac{{K}_{\mathrm{eq}}}{108{C}^{4}}={\alpha }^{5}\Rightarrow \alpha ={(\frac{{K}_{\mathrm{eq}}}{108{C}^{4}})}^{\frac{1}{5}}$

JEE Main 2023 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry