Calculate the standard enthalpy of solution of `AgCl(s)` in water `DeltaH_(f)^(0)(AgCl,s)= -127.07kJ mol^(-1), Delta H_(f)^(0)(Ag^(+),aq)=105.58 kJ mol^(-1),DeltaH_(f)^(0)(Cl^(-),aq)= -167.35 kJ mol^(-1)`

Explanation

To calculate the standard enthalpy of solution of \( \text{AgCl(s)} \) in water, we can use the following reaction:

\[ \text{AgCl(s)} \rightarrow \text{Ag}^+(aq) + \text{Cl}^-(aq) \]

The standard enthalpy of solution (\( \Delta H_{sol} \)) can be calculated using the formula:

\[ \Delta H_{sol} = \sum \Delta H_f^{\circ} \text{(products)} - \sum \Delta H_f^{\circ} \text{(reactants)} \]

### Step 1: Identify the enthalpy of formation values
We have the following values from the question:
- \( \Delta H_f^{\circ}(\text{AgCl(s)}) = -127.07 \, \text{kJ/mol} \)
- \( \Delta H_f^{\circ}(\text{Ag}^+(aq)) = 105.58 \, \text{kJ/mol} \)
- \( \Delta H_f^{\circ}(\text{Cl}^-(aq)) = -167.35 \, \text{kJ/mol} \)

### Step 2: Write the equation for \( \Delta H_{sol} \)
Using the values, we can substitute them into the equation:

\[ \Delta H_{sol} = \left( \Delta H_f^{\circ}(\text{Ag}^+) + \Delta H_f^{\circ}(\text{Cl}^-) \right) - \Delta H_f^{\circ}(\text{AgCl}) \]

### Step 3: Substitute the values into the equation
Now, substituting the values we have:

\[ \Delta H_{sol} = \left( 105.58 \, \text{kJ/mol} + (-167.35 \, \text{kJ/mol}) \right) - (-127.07 \, \text{kJ/mol}) \]

### Step 4: Calculate the sum of the products
Calculating the sum of the products:

\[ \Delta H_{sol} = \left( 105.58 - 167.35 \right) + 127.07 \]

Calculating \( 105.58 - 167.35 \):

\[ = -61.77 \, \text{kJ/mol} \]

### Step 5: Add the enthalpy of formation of \( \text{AgCl} \)
Now, adding \( 127.07 \, \text{kJ/mol} \):

\[ \Delta H_{sol} = -61.77 + 127.07 = 65.30 \, \text{kJ/mol} \]

### Final Answer
Thus, the standard enthalpy of solution of \( \text{AgCl(s)} \) in water is:

\[ \Delta H_{sol} = 65.30 \, \text{kJ/mol} \]