Question

At 298 K, the molar conductivity of $x% \ (w/w)$ MX solution (aqueous) is 123.5 S cm$^2$ mol$^{-1}$. The conductance of same solution is $1.9 \times 10^{-3}$ S. The value of $x$ is ______ $\times 10^{-2}$.
(Given : cell constant $= 1.3 \text{ cm}^{-1}$; molar mass of MX is 75 g mol$^{-1}$, density of aqueous solution of MX at 298 K is 1.0 g mL$^{-1}$)}

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
Molar conductivity ($\Lambda_m$) is related to electrolytic conductivity ($\kappa$) and molarity ($M$). We first calculate $\kappa$ using conductance and the cell constant, then find the molarity, and finally relate molarity to the mass percentage.
: Key Formula or Approach:
1. $\kappa = G \cdot \frac{l}{a}$ (Conductance $\times$ Cell Constant).
2. $\Lambda_m = \frac{\kappa \times 1000}{M}$.
3. $M = \frac{10 \times % \times d}{Mw}$.
Step 2: Detailed Explanation:
1. Calculate Conductivity ($\kappa$):
$G = 1.9 \times 10^{-3} \text{ S}$.
Cell Constant ($G^*$) = 1.3 cm$^{-1}$.
$\kappa = 1.9 \times 10^{-3} \times 1.3 = 2.47 \times 10^{-3} \text{ S cm}^{-1}$.
2. Calculate Molarity (M):
$\Lambda_m = 123.5 \text{ S cm}^2 \text{ mol}^{-1}$.
$123.5 = \frac{2.47 \times 10^{-3} \times 1000}{M} = \frac{2.47}{M}$.
$M = \frac{2.47}{123.5} = 0.02 \text{ mol/L}$.
3. Calculate Mass Percentage (x):
Density $d = 1.0 \text{ g/mL}$.
Molar mass $Mw = 75 \text{ g/mol}$.
Using $M = \frac{10 \cdot x \cdot d}{Mw}$:
$0.02 = \frac{10 \cdot x \cdot 1.0}{75}$.
$x = \frac{0.02 \times 75}{10} = \frac{1.5}{10} = 0.15$.
The question asks for the value of $x$ as $\dots \times 10^{-2}$.
$0.15 = 15 \times 10^{-2}$.
Step 3: Final Answer:
The value of $x$ is 15.