Conductivity of a solution containing one gram of anhydrous `BaCl_(2)` in 200 mL of the solution has been found to be 0.00585 `"ohm"^(-) cm^(-)`. Calculate the equivalent as well as molar conductance.

Explanation

To solve the problem step by step, we need to calculate both the molar conductance and the equivalent conductance of the solution containing one gram of anhydrous BaCl₂ in 200 mL of solution.

### Step 1: Calculate the Molecular Weight of BaCl₂
The molecular weight of BaCl₂ can be calculated by adding the atomic weights of barium (Ba) and chlorine (Cl).

- Atomic weight of Ba = 137 g/mol
- Atomic weight of Cl = 35.5 g/mol (since there are 2 Cl atoms, we multiply by 2)

\[ \text{Molecular weight of BaCl}_2 = 137 + 2 \times 35.5 = 137 + 71 = 208 \text{ g/mol} \]

### Step 2: Calculate the Molarity of the BaCl₂ Solution
Molarity (M) is defined as the number of moles of solute per liter of solution.

1. Calculate the number of moles of BaCl₂ in 1 gram:
\[ \text{Moles of BaCl}_2 = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} = \frac{1 \text{ g}}{208 \text{ g/mol}} = \frac{1}{208} \text{ mol} \]

2. Convert the volume from mL to L:
\[ \text{Volume in liters} = \frac{200 \text{ mL}}{1000} = 0.2 \text{ L} \]

3. Calculate the molarity:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{\frac{1}{208}}{0.2} = \frac{1}{208 \times 0.2} = \frac{1}{41.6} \approx 0.02404 \text{ mol/L} \]

### Step 3: Calculate the Molar Conductance (Λ_m)
Molar conductance (Λ_m) is given by the formula:
\[ \Lambda_m = \frac{K \times 1000}{\text{Molarity}} \]
where K is the conductivity.

1. Given K = 0.00585 ohm⁻¹ cm⁻¹, and using the molarity calculated:
\[ \Lambda_m = \frac{0.00585 \times 1000}{0.02404} \approx \frac{5.85}{0.02404} \approx 243.36 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1} \]

### Step 4: Calculate the Equivalent Conductance (Λ_e)
Equivalent conductance (Λ_e) is calculated using the formula:
\[ \Lambda_e = \frac{\Lambda_m}{n} \]
where n is the number of equivalents. For BaCl₂, n = 2 (since it dissociates into Ba²⁺ and 2 Cl⁻ ions).

1. Calculate Λ_e:
\[ \Lambda_e = \frac{243.36}{2} \approx 121.68 \text{ ohm}^{-1} \text{ cm}^2 \text{ eq}^{-1} \]

### Final Answers
- Molar Conductance (Λ_m) = 243.36 ohm⁻¹ cm² mol⁻¹
- Equivalent Conductance (Λ_e) = 121.68 ohm⁻¹ cm² eq⁻¹