The solubility of iodine in water is `1.1 xx 10^(-3) "mol L"^(-1)` at 288 K. When 0.200 g of iodine is stirred in 100 of water till equilibrium is reached, what will be the mass of iodine found in solution and the mass that is left undissolved. After equilibrium is reached with 0.200 g of iodine and 100 of water, we add 150 of water to the system. How much iodine will be dissolved and how much will be left undissolved and what will be the concentration of iodine in solution ?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the solubility of iodine in grams for 100 mL of water
The solubility of iodine in water is given as \(1.1 \times 10^{-3} \, \text{mol L}^{-1}\) at 288 K.

1. **Convert the solubility to grams**:
- The molar mass of iodine (I₂) is approximately 254 g/mol.
- In 1 L (1000 mL) of water, the amount of iodine that can dissolve is:
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