JEE Main 2026 — Electrochemistry Question with Solution

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

$\kappa =G\times G^{\ast }$

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

The molar conductivity, ${\Lambda }_m$, is related to the conductivity and molarity $C$ by the formula:

${\Lambda }_m=\frac{\kappa \times 1000}{C}$

Substituting the given values:

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

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

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

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

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

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

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

$1.5=10x$

$x=0.15$

Expressing $x$ in the required format:

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

The value is $15$.

Answer: $15$