Q24 / 75 · JEE Main 2026 (05 April Shift 2) · Chemistry
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 $1.0\text{ g mL}^{-1}$)
(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 $1.0\text{ g mL}^{-1}$)
Correct answer: 15
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: $15$
$\kappa = G \times G^*$
$\kappa = 1.9 \times 10^{-3} \text{ S} \times 1.3 \text{ cm}^{-1} = 2.47 \times 10^{-3} \text{ S cm}^{-1}$
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: $15$
Hence, the correct answer is 15.