The $$K_{sp}$$ for bismuth sulphide $$Bi_2S_3$$ is $$1.08 \times 10^{-73}$$. The solubility of $$Bi_2S_3$$ in mol L$$^{-1}$$ at $$298$$ K is

$$K_{sp}$$ of $$Bi_2S_3 = 1.08 \times 10^{-73}$$

$$Bi_2S_3(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$$

Let the solubility of $$Bi_2S_3$$ be $$s$$ mol/L.

Then: $$[Bi^{3+}] = 2s$$ and $$[S^{2-}] = 3s$$

$$K_{sp} = [Bi^{3+}]^2 [S^{2-}]^3 = (2s)^2(3s)^3$$

$$= 4s^2 \times 27s^3 = 108s^5$$

$$108s^5 = 1.08 \times 10^{-73}$$

$$s^5 = \frac{1.08 \times 10^{-73}}{108} = 1.0 \times 10^{-75}$$

$$s = (10^{-75})^{1/5} = 10^{-15}$$

$$s = 1.0 \times 10^{-15} \text{ mol/L}$$

Hence, the correct answer is Option A.