Calculate the magnitude of free energy in `KJ mol^(-1)` when `1` mole of a an ionic salt MX (s) is dissolved in water at `27^(@)C` . Given
Lattice energy of MX `= 780 KJ mol^(-1)`
Hydration energy of MX `=-775.0 KJ mol^(-1)`
Entropy change of dissolution at `27^(@)C = 40Jmol^(-1)K^(-1)`

Explanation

To calculate the magnitude of free energy (ΔG) when 1 mole of an ionic salt MX(s) is dissolved in water at 27°C, we will use the following information:

- Lattice energy of MX, ΔH_lattice = 780 kJ/mol
- Hydration energy of MX, ΔH_hydration = -775.0 kJ/mol
- Entropy change of dissolution, ΔS = 40 J/mol·K
- Temperature, T = 27°C = 300 K (since T in Kelvin = T in °C + 273)

### Step-by-Step Solution:

**Step 1: Calculate the Enthalpy Change of Dissolution (ΔH_dissolution)**

The enthalpy change of dissolution can be calculated using the formula:
\[ \Delta H_{dissolution} = \Delta H_{lattice} + \Delta H_{hydration} \]

Substituting the values:
\[ \Delta H_{dissolution} = 780 \, \text{kJ/mol} + (-775.0 \, \text{kJ/mol}) \]
\[ \Delta H_{dissolution} = 780 - 775 = 5 \, \text{kJ/mol} \]

**Step 2: Convert ΔH_dissolution to Joules**

Since ΔS is given in J/mol·K, we need to convert ΔH_dissolution to Joules:
\[ \Delta H_{dissolution} = 5 \, \text{kJ/mol} = 5000 \, \text{J/mol} \]

**Step 3: Calculate the Gibbs Free Energy Change (ΔG)**

The Gibbs free energy change can be calculated using the formula:
\[ \Delta G = \Delta H_{dissolution} - T \Delta S \]

Substituting the values:
\[ \Delta G = 5000 \, \text{J/mol} - (300 \, \text{K} \times 40 \, \text{J/mol·K}) \]
\[ \Delta G = 5000 \, \text{J/mol} - 12000 \, \text{J/mol} \]
\[ \Delta G = 5000 - 12000 = -7000 \, \text{J/mol} \]

**Step 4: Convert ΔG to kJ/mol**

Finally, convert ΔG back to kJ/mol:
\[ \Delta G = -7000 \, \text{J/mol} = -7 \, \text{kJ/mol} \]

### Final Answer:
The magnitude of free energy change when 1 mole of ionic salt MX(s) is dissolved in water at 27°C is:
\[ \Delta G = -7 \, \text{kJ/mol} \]