JEE Main 2026 — Chemistry Physical Chemistry

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At $298\text{ K}$ , the molar conductivity of $x\%$ (w/w) $MX$ solution (aqueous) is $123.5\text{ S cm}^2\text{ mol}^{-1}$ . The conductance of same solution is $1.9 \times 10^{-3}\text{ S}$ . The value of $x$ is _______ $\times 10^{-2}$ .

(Given : cell constant $= 1.3\text{ cm}^{-1}$ ; molar mass of $MX$ is $75\text{ g mol}^{-1}$ , density of aqueous solution of $MX$ at $298\text{ K}$ is $1.0\text{ g mL}^{-1}$ )

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

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Solution

The conductivity (specific conductance), $\kappa$ , is given by the product of conductance and cell constant:

$\kappa = G \times G^*$

$\kappa = 1.9 \times 10^{-3} \text{ S} \times 1.3 \text{ cm}^{-1} = 2.47 \times 10^{-3} \text{ S cm}^{-1}$

The molar conductivity, $\Lambda_m$ , is related to the conductivity and molarity $C$ by the formula:

$\Lambda_m = \dfrac{\kappa \times 1000}{C}$

Substituting the given values:

$123.5 = \dfrac{2.47 \times 10^{-3} \times 1000}{C}$

$123.5 = \dfrac{2.47}{C}$

$C = \dfrac{2.47}{123.5} = 0.02 \text{ mol L}^{-1}$

Molarity $C$ can also be expressed in terms of mass percentage $x$ (w/w), density $d$ , and molar mass $M$ :

$C = \dfrac{x \times d \times 10}{M}$

Substituting the known values ( $d = 1.0 \text{ g mL}^{-1}$ , $M = 75 \text{ g mol}^{-1}$ ):

$0.02 = \dfrac{x \times 1.0 \times 10}{75}$

$1.5 = 10x$

$x = 0.15$

Expressing $x$ in the required format:

$x = 15 \times 10^{-2}$

The value is $15$ .

Answer: $15$

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Solution

The conductivity (specific conductance), $\kappa$ , is given by the product of conductance and cell constant:

$\kappa = G \times G^*$

$\kappa = 1.9 \times 10^{-3} \text{ S} \times 1.3 \text{ cm}^{-1} = 2.47 \times 10^{-3} \text{ S cm}^{-1}$

The molar conductivity, $\Lambda_m$ , is related to the conductivity and molarity $C$ by the formula:

$\Lambda_m = \dfrac{\kappa \times 1000}{C}$

Substituting the given values:

$123.5 = \dfrac{2.47 \times 10^{-3} \times 1000}{C}$

$123.5 = \dfrac{2.47}{C}$

$C = \dfrac{2.47}{123.5} = 0.02 \text{ mol L}^{-1}$

Molarity $C$ can also be expressed in terms of mass percentage $x$ (w/w), density $d$ , and molar mass $M$ :

$C = \dfrac{x \times d \times 10}{M}$

Substituting the known values ( $d = 1.0 \text{ g mL}^{-1}$ , $M = 75 \text{ g mol}^{-1}$ ):

$0.02 = \dfrac{x \times 1.0 \times 10}{75}$

$1.5 = 10x$

$x = 0.15$

Expressing $x$ in the required format:

$x = 15 \times 10^{-2}$

The value is $15$ .

Answer: $15$

JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

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Organic Chemistry

Inorganic Chemistry

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