To determine the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution, let’s consider the solubility equilibrium and how it is affected by the common ion effect.

First, let’s establish the dissociation equation for Ag₂CrO₄ :

$$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq)$$

The solubility product constant, K_sp, is given by :

$$K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$$

Given that K_sp = 1.1 $$\times$$ 10^-12.

In a 0.1 M AgNO₃ solution, the concentration of Ag⁺ ions from AgNO₃ alone is already 0.1 M. Let’s denote the additional solubility of Ag₂CrO₄ in this solution as $s$.

Therefore, the total concentration of Ag⁺ ions will be (0.1 + 2s) M, and the concentration of CrO₄^2- ions will be $s$ M.

Substituting these into the K_sp expression :

$$K_{sp} = (0.1 + 2s)^2 s$$

Given that K_sp is a very small number, we can assume that $2s$ is much smaller than 0.1, therefore $0.1 + 2s \approx 0.1$. This simplifies our equation to :

$$K_{sp} \approx (0.1)^2 s$$

Substitute $K_{sp} = 1.1 \times 10^{-12}$ into the equation :

$$1.1 \times 10^{-12} \approx (0.1)^2 s$$

$$1.1 \times 10^{-12} \approx 0.01 s$$

Solving for $s$ :

$$s = \frac{1.1 \times 10^{-12}}{0.01} = 1.1 \times 10^{-10}$$

Hence, the solubility of Ag₂CrO₄ in a 0.1 M AgNO₃ solution is 1.1 $$\times$$ 10^-10 mol/L.

The correct option is B: 1.1 $$\times$$ 10^-10.