Question:
Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at \(25^{\circ}\)C are respectively 120 ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\) and 0.0024 ohm\(^{-1}\) cm\(^{-1}\).
Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at \(25^{\circ}\)C are respectively 120 ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\) and 0.0024 ohm\(^{-1}\) cm\(^{-1}\).
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Molar conductivity = (conductivity x 1000) / concentration.
Updated On: Oct 1, 2026
- \(0.01\) M
- \(0.02\) M
- \(0.03\) M
- \(0.04\) M
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Molar conductivity \(\Lambda_m\) relates the conductivity \(\kappa\) of a solution to its molar concentration \(c\) (in mol/L).
Step 2: Key Formula or Approach
\[ \Lambda_m = \frac{\kappa \times 1000}{c} \]
Step 3: Detailed Explanation
\[ c = \frac{\kappa \times 1000}{\Lambda_m} = \frac{0.0024 \times 1000}{120} = \frac{2.4}{120} = 0.02 \text{ M} \]
The factor 1000 converts \(\text{cm}^3\) to litres. The other options would come from wrongly using 1000 or dividing by a different value.
Final Answer:
The concentration is 0.02 M, option (B). \[ \boxed{0.02 \text{ M}} \]
Molar conductivity \(\Lambda_m\) relates the conductivity \(\kappa\) of a solution to its molar concentration \(c\) (in mol/L).
Step 2: Key Formula or Approach
\[ \Lambda_m = \frac{\kappa \times 1000}{c} \]
Step 3: Detailed Explanation
\[ c = \frac{\kappa \times 1000}{\Lambda_m} = \frac{0.0024 \times 1000}{120} = \frac{2.4}{120} = 0.02 \text{ M} \]
The factor 1000 converts \(\text{cm}^3\) to litres. The other options would come from wrongly using 1000 or dividing by a different value.
Final Answer:
The concentration is 0.02 M, option (B). \[ \boxed{0.02 \text{ M}} \]
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