x g of Ag was dissolved in `HNO_(3)` and the solution was treated with excess NaCl when 2.87 gm. of AgCl was precipitated precipitated. The value of x is

Options

A. `1*08 g`
B. `2*16 g`
C. `2*70 g`
D. `1*62 g`

Explanation

To find the value of x (the mass of silver, Ag, that was dissolved), we can follow these steps:

### Step 1: Write the reaction
When silver (Ag) reacts with sodium chloride (NaCl), silver chloride (AgCl) is formed. The reaction can be represented as:
\[ \text{Ag} + \text{NaCl} \rightarrow \text{AgCl} + \text{Na} \]

### Step 2: Determine the molar masses
- Molar mass of Ag = 108 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of AgCl = 108 g/mol + 35.5 g/mol = 143.5 g/mol

### Step 3: Set up the proportion
From the reaction, we know that:
- 108 g of Ag produces 143.5 g of AgCl.

We can set up a proportion to find out how much Ag is needed to produce 2.87 g of AgCl:
\[ \frac{108 \text{ g Ag}}{143.5 \text{ g AgCl}} = \frac{x \text{ g Ag}}{2.87 \text{ g AgCl}} \]

### Step 4: Cross-multiply and solve for x
Cross-multiplying gives us:
\[ x \cdot 143.5 = 108 \cdot 2.87 \]
Now, calculate the right side:
\[ x \cdot 143.5 = 309.96 \]
Now, divide both sides by 143.5 to isolate x:
\[ x = \frac{309.96}{143.5} \approx 2.16 \text{ g} \]

### Conclusion
The value of x, which represents the mass of silver that was dissolved, is approximately **2.16 g**.

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