In a sautrated solution of the sparingly soluble strong electrolyte $$AgIO_3$$ (Molecular mass = 283) the equilibrium which sets in is $$AgIO_{3(s)} \rightleftharpoons Ag^+_{(aq)} + IO^-_{3(aq)}$$. If the solubility product constant $$K_{sp}$$ of $$AgIO_3$$ at a given temperature is $$1.0 \times 10^{-8}$$, what is the mass of $$AgIO_3$$ contained in 100 ml of its saturated solution?

The salt $$AgIO_3$$ is sparingly soluble and dissociates as
$$AgIO_{3(s)} \rightleftharpoons Ag^+_{(aq)} + IO_3^-{}_{(aq)}$$

If the molar solubility of the salt is $$S$$ mol L$$^{-1}$$, then at equilibrium
$$\left[Ag^+\right] = S \quad\text{and}\quad \left[IO_3^-\right] = S$$

The solubility-product expression is
$$K_{sp} = [Ag^+][IO_3^-] = S \times S = S^2$$

Given $$K_{sp} = 1.0 \times 10^{-8}$$, we get
$$S^2 = 1.0 \times 10^{-8} \;\;\Longrightarrow\;\; S = 1.0 \times 10^{-4}\,\text{mol L}^{-1}$$

The volume of solution considered is $$100\ \text{mL} = 0.10\ \text{L}$$, so the number of moles of $$AgIO_3$$ present is
$$(0.10\ \text{L}) \times (1.0 \times 10^{-4}\ \text{mol L}^{-1}) = 1.0 \times 10^{-5}\ \text{mol}$$

Using the molar mass $$M = 283\ \text{g mol}^{-1}$$,
$$\text{mass} = nM = (1.0 \times 10^{-5}\ \text{mol}) \times (283\ \text{g mol}^{-1}) = 2.83 \times 10^{-3}\ \text{g}$$

Hence, the mass of $$AgIO_3$$ in 100 mL of its saturated solution is $$2.83 \times 10^{-3}$$ g.
Option B which is: $$2.83 \times 10^{-3}$$ g