If the solubility product of PbS is $$8 \times 10^{-28}$$, then the solubility of PbS in pure water at 298 K is $$x \times 10^{-16}$$ mol L$$^{-1}$$. The value of x is _____ (Nearest integer) [Given $$\sqrt{2} = 1.41$$]

We are given the solubility product of PbS as $$K_{sp} = 8 \times 10^{-28}$$ and need to find its solubility in pure water.

PbS dissociates in water as:

$$PbS \rightleftharpoons Pb^{2+} + S^{2-}$$

Let the solubility be $$s$$ mol L$$^{-1}$$. Then $$[Pb^{2+}] = s$$ and $$[S^{2-}] = s$$.

The solubility product expression is:

$$K_{sp} = [Pb^{2+}][S^{2-}] = s \times s = s^2$$

$$s^2 = 8 \times 10^{-28}$$

$$s = \sqrt{8 \times 10^{-28}} = \sqrt{8} \times 10^{-14}$$

Now, $$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} = 2 \times 1.41 = 2.82$$.

$$s = 2.82 \times 10^{-14} \text{ mol L}^{-1}$$

We need to express this as $$x \times 10^{-16}$$ mol L$$^{-1}$$:

$$s = 2.82 \times 10^{-14} = 282 \times 10^{-16} \text{ mol L}^{-1}$$

Hence, the correct answer is 282.