Question

The molar conductivity of \(0.5 \, \text{mol/dm}^3\) solution of \(AgNO_3\) with electrolytic conductivity of \(5.76 \times 10^{-3} \, \text{S cm}^{-1}\) at \(298\,K\) is:

• A. \(0.086 \, \text{S cm}^2\text{/mol}\)
• B. \(28.8 \, \text{S cm}^2\text{/mol}\)
• C. \(2.88 \, \text{S cm}^2\text{/mol}\)
• D. \(11.52 \, \text{S cm}^2\text{/mol}\)

Hint:

While calculating molar conductivity, always remember the formula: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where concentration must be in \(\text{mol L}^{-1}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The question asks for the molar conductivity ($\Lambda_m$) of a silver nitrate ($AgNO_3$) solution.
Molar conductivity is the conductivity of an electrolytic solution divided by the molar concentration of the electrolyte.
It represents the efficiency with which a given electrolyte conducts electricity in solution.
We are given the electrolytic conductivity ($\kappa$), which is the conductance of a solution of 1 cm length and 1 cm$^2$ cross-section area.
The concentration is given in $mol/dm^3$, which is equivalent to Molarity ($M$).
Units are critical here; conductivity is in $S \cdot cm^{-1}$, so we must ensure the volume units are consistent ($1 L = 1000 cm^3$).
Step 2: Key Formula or Approach:
The standard formula relating molar conductivity ($\Lambda_m$), electrolytic conductivity ($\kappa$), and molarity ($M$) is:
\[ \Lambda_m = \frac{\kappa \times 1000}{M} \]
Where:
$\kappa$ = Electrolytic conductivity (Specific conductance) in $S \cdot cm^{-1}$.
$M$ = Molarity of the solution in $mol \cdot L^{-1}$ (or $mol/dm^3$).
1000 = Conversion factor to convert $dm^3$ (liters) into $cm^3$.
Step 3: Detailed Explanation:
First, identify the given values:
$\kappa = 5.76 \times 10^{-3} \text{ S cm}^{-1}$
$M = 0.5 \text{ mol/dm}^3 = 0.5 \text{ mol/L}$
Temperature = $298 K$ (This is the standard temperature for measurement but does not change the calculation).

Substitute the values into the molar conductivity formula:
\[ \Lambda_m = \frac{5.76 \times 10^{-3} \times 1000}{0.5} \]

Calculate the numerator:
\[ 5.76 \times 10^{-3} \times 10^3 = 5.76 \]

Now, divide by the concentration:
\[ \Lambda_m = \frac{5.76}{0.5} \]
Dividing by 0.5 is mathematically the same as multiplying by 2.
\[ \Lambda_m = 11.52 \text{ S cm}^2 \text{ mol}^{-1} \]

The units match the requirements for option (D).
$AgNO_3$ is a strong electrolyte, meaning it dissociates completely into $Ag^+$ and $NO_3^-$ ions.
At a concentration of $0.5 M$, the ions are close enough to exert inter-ionic attractions, which is why the molar conductivity is less than the limiting molar conductivity ($\Lambda_m^\circ$) found at infinite dilution.
However, for this specific calculation, the formula provided leads directly to the result.
Step 4: Final Answer:
The molar conductivity is calculated as 11.52 S cm$^2$/mol.