`2.66 g` of chloride of a metal when treated with silver nitrate solution, gave `2.87 g` of silver chloride `3.37 g` of another chloride of the same metal gave `5.74 g` of silver chloride when treated with silver nitrate solution. Show that the results illustrate the law of multiple proportions.

Explanation

To demonstrate that the results illustrate the law of multiple proportions, we will follow these steps:

### Step 1: Determine the mass of chlorine in each chloride.
When a metal chloride reacts with silver nitrate, silver chloride (AgCl) is formed. The mass of silver chloride produced can be used to find the mass of chlorine in the metal chloride.

1. **For the first chloride:**
- Mass of silver chloride = 2.87 g
- Molar mass of AgCl = 143.32 g/mol (Ag = 107.87 g/mol, Cl = 35.45 g/mol)
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