For the following reaction
`Ag_((aq))^(+)+Cl_((aq))^(-)rarrAgCl_((s))`
Given : `DeltaG_(f)^(@), AgCl=-"112.44 kJ/mol," DeltaG_(f)^(@) Cl^(-)=-"130 kJ/mol", DeltaG_(f)^(@)Ag^(+)="75 kJ/mol"` Report your answer by rounding it upto nearest whole number. The `K_(sp)` of `AgCl` is `nxx10^(-10)`. The value of 'n'' is .

Answer

To solve the given problem, we need to calculate the solubility product constant (Ksp) for the reaction:

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

We are provided with the following standard Gibbs free energies of formation:

- \( \Delta G_f^\circ (\text{AgCl}) = -112.44 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Cl}^-) = -130 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ (\text{Ag}^+) = 75 \, \text{kJ/mol} \)

### Step 1: Calculate \( \Delta G^\circ \) for the reaction

The standard Gibbs free energy change for the reaction can be calculated using the formula:

\[ \Delta G^\circ = \Delta G_f^\circ (\text{products}) - \Delta G_f^\circ (\text{reactants}) \]

Substituting the values:

\[ \Delta G^\circ = \Delta G_f^\circ (\text{AgCl}) - \left( \Delta G_f^\circ (\text{Ag}^+) + \Delta G_f^\circ (\text{Cl}^-) \right) \]

\[ \Delta G^\circ = -112.44 - (75 - 130) \]

\[ \Delta G^\circ = -112.44 - (75 - 130) = -112.44 - 75 + 130 \]

\[ \Delta G^\circ = -112.44 - 75 + 130 = -57.44 \, \text{kJ/mol} \]

### Step 2: Convert \( \Delta G^\circ \) to Joules

Since we need to work in Joules for the next calculations, we convert \( \Delta G^\circ \):

\[ \Delta G^\circ = -57.44 \, \text{kJ/mol} \times 1000 = -57440 \, \text{J/mol} \]

### Step 3: Use the relationship between \( \Delta G^\circ \) and \( K_{sp} \)

The relationship between Gibbs free energy and the solubility product constant is given by:

\[ \Delta G^\circ = -RT \ln K_{sp} \]

Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 300 \, \text{K} \) (assuming room temperature)

Rearranging gives:

\[ \ln K_{sp} = -\frac{\Delta G^\circ}{RT} \]

Substituting the values:

\[ \ln K_{sp} = -\frac{-57440}{8.314 \times 300} \]

Calculating the right side:

\[ \ln K_{sp} = \frac{57440}{2494.2} \approx 23.0 \]

### Step 4: Calculate \( K_{sp} \)

To find \( K_{sp} \), we take the exponential:

\[ K_{sp} = e^{23.0} \]

Calculating \( K_{sp} \):

\[ K_{sp} \approx 9.74 \times 10^{9} \]

### Step 5: Express \( K_{sp} \) in the form \( n \times 10^{-10} \)

To express \( K_{sp} \) in the form \( n \times 10^{-10} \):

\[ K_{sp} = 9.74 \times 10^{9} = 97.4 \times 10^{8} = 0.974 \times 10^{10} \]

Thus, rounding \( n \) to the nearest whole number gives:

\[ n = 97 \]

### Final Answer

The value of \( n \) is \( 97 \).