At 298 K , the molar conductivity of x\% (w/w) MX 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: $15$