At $$310 \text{ K}$$, the solubility of $$CaF_2$$ in water is $$2.34 \times 10^{-3} \text{ g/100 mL}$$. The solubility product of $$CaF_2$$ is ______ $$\times 10^{-8} (\text{mol/L})^3$$ (nearest integer). (Given molar mass: $$CaF_2 = 78 \text{ g mol}^{-1}$$)

We need to find the solubility product $$K_{sp}$$ of $$CaF_2$$.

Since the solubility of $$CaF_2$$ at 310 K is $$2.34 \times 10^{-3} \text{ g/100 mL}$$ and the molar mass of $$CaF_2$$ is $$78 \text{ g mol}^{-1}$$, we first convert the solubility to mol/L.

Because the solubility is given as $$2.34 \times 10^{-3} \text{ g per 100 mL}$$, converting to grams per liter involves multiplying by 10, which yields $$2.34 \times 10^{-2} \text{ g/L}$$. Substituting this into the molar solubility formula gives $$s = \frac{2.34 \times 10^{-2}}{78} = 3.0 \times 10^{-4} \text{ mol/L}$$.

Next, we write the dissociation equation: $$CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$$. If $$s$$ is the molar solubility, then $$[Ca^{2+}] = s = 3.0 \times 10^{-4} \text{ M}$$ and $$[F^-] = 2s = 6.0 \times 10^{-4} \text{ M}$$.

To calculate $$K_{sp}$$, we use the expression $$K_{sp} = [Ca^{2+}][F^-]^2 = s \times (2s)^2 = 4s^3$$. Substituting the value of $$s$$ gives $$K_{sp} = 4 \times (3.0 \times 10^{-4})^3$$, which simplifies to $$K_{sp} = 4 \times 2.7 \times 10^{-11}$$, then $$K_{sp} = 10.8 \times 10^{-11}$$, and finally $$K_{sp} = 1.08 \times 10^{-10}$$.

Expressing this in the required form yields $$K_{sp} = 1.08 \times 10^{-10} = 0.0108 \times 10^{-8} \approx 0 \times 10^{-8}$$. Rounding to the nearest integer in units of $$\times 10^{-8} (\text{mol/L})^3$$ gives $$K_{sp} = 0.0108 \times 10^{-8}$$.

The nearest integer is 0, and therefore the correct answer is 0.