The resistance of an aqueous solution containing `0.624g` of `CuSO_(4).5H_(2)O` per `100 cm^(3)` of the solution in a conductance cell of cell constant `153.7` per meter iss 520 ohms at 298K. Calculate the moalr conductivity. `(CuSO_(4).5H_(2)O = 249.5)`

Explanation

To calculate the molar conductivity of the aqueous solution of CuSO₄·5H₂O, we will follow these steps:

### Step 1: Calculate the specific conductivity (κ)
The specific conductivity (κ) can be calculated using the formula:
\[ \kappa = \frac{1}{R} \times G^* \]
where:
- \( R \) is the resistance of the solution (520 ohms),
- \( G^* \) is the cell constant (153.7 m⁻¹).

Substituting the values:
\[ \kappa = \frac{1}{520} \times 153.7 \]
Calculating this gives:
\[ \kappa = \frac{153.7}{520} \approx 0.295 \, \text{S/m} \text{ (or } 2.95 \times 10^{-1} \text{ S/m)} \]

### Step 2: Calculate the concentration (C) of the solution
To find the concentration, we first need to calculate the number of moles of CuSO₄·5H₂O in the solution. The molar mass of CuSO₄·5H₂O is given as 249.5 g/mol.

First, calculate the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.624 \, \text{g}}{249.5 \, \text{g/mol}} \approx 0.0025 \, \text{mol} \]

Now, since the volume of the solution is 100 cm³ (which is 0.1 L), the concentration (C) in mol/L is:
\[ C = \frac{\text{moles}}{\text{volume}} = \frac{0.0025 \, \text{mol}}{0.1 \, \text{L}} = 0.025 \, \text{mol/L} \]

### Step 3: Calculate the molar conductivity (Λm)
The molar conductivity (Λm) can be calculated using the formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
Substituting the values we have:
\[ \Lambda_m = \frac{0.295 \times 1000}{0.025} \]
Calculating this gives:
\[ \Lambda_m = \frac{295}{0.025} = 11800 \, \text{S cm²/mol} \]

### Final Result
Thus, the molar conductivity of the solution is:
\[ \Lambda_m \approx 11800 \, \text{S cm²/mol} \]