Solubility of iodine of water is `1.1 times 10^-3 mol L^-1` at 290K. When 0.1g of iodine is stirred in 200ml water till the equilibrium is reached at 290K, calculate the mass of undissolved iodine.

Explanation

To solve the problem, we need to calculate the mass of undissolved iodine after reaching equilibrium in water. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert the solubility of iodine from moles per liter to grams per liter.
The solubility of iodine in water is given as \(1.1 \times 10^{-3} \, \text{mol L}^{-1}\). To convert this to grams per liter, we need to multiply by the molecular weight of iodine.

**Molecular weight of iodine (I₂)**:
- The molecular weight of iodine (I₂) is approximately 254 g/mol.

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