JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Given is a concentrated solution of a weak electrolyte $A_xB_y$ of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\left[K \times c^{x+y-1} x^x y^y\right]^{x+y}$
B
$\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$
C
$\left(\dfrac{c^{x+y-1} x^x y^y}{K}\right)^{x+y}$
D
$\left(\dfrac{c^{x+y-1} x^x y^y}{K}\right)^{\frac{1}{x+y}}$

Did you get this right?

Solution

The dissociation reaction of the weak electrolyte $A_xB_y$ is:

$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$

Initial concentrations:

$c \quad 0 \quad 0$

Equilibrium concentrations:

$c(1-\alpha) \quad cx\alpha \quad cy\alpha$

The dissociation constant $K$ is given by:

$K = \dfrac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]}$

$K = \dfrac{(cx\alpha)^x (cy\alpha)^y}{c(1-\alpha)}$

$K = \dfrac{c^{x+y} x^x y^y \alpha^{x+y}}{c(1-\alpha)}$

$K = \dfrac{c^{x+y-1} x^x y^y \alpha^{x+y}}{1-\alpha}$

For a weak electrolyte, the degree of dissociation $\alpha$ is very small, so $1-\alpha \approx 1$ .

$K = c^{x+y-1} x^x y^y \alpha^{x+y}$

Solving for $\alpha$ :

$\alpha^{x+y} = \dfrac{K}{c^{x+y-1} x^x y^y}$

$\alpha = \left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\dfrac{1}{x+y}}$

Answer: $\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Given is a concentrated solution of a weak electrolyte $A_xB_y$ of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\left[K \times c^{x+y-1} x^x y^y\right]^{x+y}$
B
$\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$
C
$\left(\dfrac{c^{x+y-1} x^x y^y}{K}\right)^{x+y}$
D
$\left(\dfrac{c^{x+y-1} x^x y^y}{K}\right)^{\frac{1}{x+y}}$

Did you get this right?

Solution

The dissociation reaction of the weak electrolyte $A_xB_y$ is:

$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$

Initial concentrations:

$c \quad 0 \quad 0$

Equilibrium concentrations:

$c(1-\alpha) \quad cx\alpha \quad cy\alpha$

The dissociation constant $K$ is given by:

$K = \dfrac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]}$

$K = \dfrac{(cx\alpha)^x (cy\alpha)^y}{c(1-\alpha)}$

$K = \dfrac{c^{x+y} x^x y^y \alpha^{x+y}}{c(1-\alpha)}$

$K = \dfrac{c^{x+y-1} x^x y^y \alpha^{x+y}}{1-\alpha}$

For a weak electrolyte, the degree of dissociation $\alpha$ is very small, so $1-\alpha \approx 1$ .

$K = c^{x+y-1} x^x y^y \alpha^{x+y}$

Solving for $\alpha$ :

$\alpha^{x+y} = \dfrac{K}{c^{x+y-1} x^x y^y}$

$\alpha = \left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\dfrac{1}{x+y}}$

Answer: $\left(\dfrac{K}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2026 — 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 2026 — 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