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Step-by-Step Solution
Step 1: Identify the Given Data for 0.2 M Solution
• Concentration: 0.2 M
• Resistance, R₁ = 50 Ω
• Specific conductance (κ₁) = 1.4 S m⁻¹
• Convert κ₁ from S m⁻¹ to S cm⁻¹:
$$
1\text{ S m}^{-1} = 10^{-2}\text{ S cm}^{-1} \\
\therefore \kappa_{1} = 1.4 \times 10^{-2}\text{ S cm}^{-1}
$$
Step 2: Calculate the Cell Constant (ℓ/a)
The cell constant, $\frac{\ell}{a}$, can be found using the relation:
$$
R = \rho \frac{\ell}{a} = \frac{1}{\kappa}\,\frac{\ell}{a} \quad\Longrightarrow\quad \frac{\ell}{a} = R \times \kappa
$$
Substituting R₁ = 50 Ω and κ₁ = 1.4 × 10⁻² S cm⁻¹:
$$
\frac{\ell}{a} = 50 \times 1.4 \times 10^{-2} = 0.7 \text{ (dimension consistent for the cell constant)}
$$
Step 3: Identify the Given Data for 0.5 M Solution
• Concentration: 0.5 M
• Resistance, R₂ = 280 Ω
• Specific conductance (κ₂) = ? (to be found)
Step 4: Calculate the Specific Conductance (κ₂) for 0.5 M
Using the same cell constant (ℓ/a = 0.7), we write:
$$
R = \frac{1}{\kappa}\,\frac{\ell}{a} \quad\Longrightarrow\quad \kappa = \frac{1}{R} \times \frac{\ell}{a}.
$$
Substituting R₂ = 280 Ω and ℓ/a = 0.7:
$$
\kappa_2 = \frac{1}{280} \times 0.7 = 2.5 \times 10^{-3}\,\text{S cm}^{-1}.
$$
Step 5: Calculate the Molar Conductivity (Λₘ) for 0.5 M
The formula for molar conductivity is:
$$
\Lambda_{m} = \frac{\kappa \times 1000}{c}
$$
where:
• $ \kappa $ is in S cm⁻¹
• c is the concentration in mol cm⁻³ (generally we use mol L⁻¹, so 1000 cm³ = 1 L)
• The resulting $\Lambda_{m}$ is in S cm² mol⁻¹
Substituting $\kappa_2 = 2.5 \times 10^{-3}\,\text{S cm}^{-1}$ and c = 0.5 M:
$$
\Lambda_{m} = \frac{2.5 \times 10^{-3} \times 1000}{0.5} = 5\,\text{S cm}^2 \text{mol}^{-1}.
$$
Step 6: Convert to SI Units (S m² mol⁻¹)
1 cm = 10⁻² m, hence 1 cm² = 10⁻⁴ m². To convert S cm² mol⁻¹ to S m² mol⁻¹, we multiply by 10⁻⁴:
$$
5\,\text{S cm}^2 \text{mol}^{-1} = 5 \times 10^{-4}\,\text{S m}^2 \text{mol}^{-1}.
$$
Final Answer
The molar conductivity of the 0.5 M solution is
$5 \times 10^{-4}\,\text{S m}^2 \text{mol}^{-1}$.
