Q5. If 9.62 g of potassium iodide is dissolved in sufficient water to make a 4.50 L solution, then what is its molarity? (2 marks)

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1. Calculate the number of moles of potassium iodide using the formula
   $$
   \text{moles} = \frac{\text{mass}}{\text{molar mass}}.
   $$
   The molar mass of potassium iodide ($$ \text{KI} $$) is
   $$
   M_{\text{KI}} = M_{\text{K}} + M_{\text{I}} = 39.10\,\text{g/mol} + 126.90\,\text{g/mol} = 166.00\,\text{g/mol}.
   $$
   Thus, the number of moles is
   $$
   n = \frac{9.62\,\text{g}}{166.00\,\text{g/mol}} \approx 0.05795\,\text{mol}.
   $$

2. Calculate the molarity using the definition
   $$
   \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}.
   $$
   Given a volume of $$ 4.50\,\text{L} $$, the molarity ($$ M $$) is
   $$
   M = \frac{0.05795\,\text{mol}}{4.50\,\text{L}} \approx 0.01288\,\text{M}.
   $$

The molarity of the solution is approximately $$ 0.0129\,\text{M} $$.
The molarity of the potassium iodide solution is approximately 0.0129 M.

Q5. If 9.62 g of potassium iodide is dissolved in sufficient water to make a 4.50 L solution, then what is its molarity? (2 marks)

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