If `6.65xx10^(-2)g` of metalllic zinc is added to `100ml` saturated solution of `AgCl`, it react with `Ag^(+)` of solution as following reaction.
`Zn(s)+2Ag^(+)(aq)hArrZn^(2+)(aq)+2Ag(s)`
and approximately `10^(-x)` moles of `Ag` will be precipated. Calculate the value of `x` (Given `E_(Zn^(++)//Zn)^(@)=-0.76VE_(Ag^(+)//Ag)^(@)=0.8V, K_(sp)` of `AgCl=10^(-10)`, atomic mass of `Zn=65.3u, 10^(52.8813)=7.61xx10^(52)`)

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the given information.

### Step 1: Calculate the number of moles of Zinc (Zn)

Given:
- Mass of Zinc = \(6.65 \times 10^{-2} \, \text{g}\)
- Atomic mass of Zinc = \(65.3 \, \text{g/mol}\)

Using the formula for moles:
\[ \text{Moles of Zn} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{6.65 \times 10^{-2}}{65.3} \]

Calculating:
\[ \text{Moles of Zn} \approx \frac{6.65 \times 10^{-2}}{65.3} \approx 1.018 \times 10^{-3} \, \text{mol} \]

### Step 2: Determine the moles of Silver (Ag) produced

From the balanced reaction:
\[ \text{Zn} + 2 \text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2 \text{Ag} \]

This indicates that 1 mole of Zn produces 2 moles of Ag. Therefore, the moles of Ag produced will be:
\[ \text{Moles of Ag} = 2 \times \text{Moles of Zn} = 2 \times 1.018 \times 10^{-3} \approx 2.036 \times 10^{-3} \, \text{mol} \]

### Step 3: Calculate the value of \(x\)

According to the problem, the number of moles of Ag precipitated is given as approximately \(10^{-x}\) moles. We have calculated:
\[ \text{Moles of Ag} \approx 2.036 \times 10^{-3} \, \text{mol} \approx 2.0 \times 10^{-3} \, \text{mol} \]

This can be expressed as:
\[ 2.0 \times 10^{-3} = 10^{-3} \times 2 \]

Since \(10^{-3}\) is the dominant term, we can approximate:
\[ 10^{-x} \approx 10^{-3} \implies x = 3 \]

### Conclusion

Thus, the value of \(x\) is:
\[ \boxed{3} \]