The heat of solution of anhydrous $$CuSO_4$$ and $$CuSO_4 \cdot 5H_2O$$ are $$-70$$ kJ mol$$^{-1}$$ and $$+12$$ kJ mol$$^{-1}$$ respectively. The heat of hydration of $$CuSO_4$$ to $$CuSO_4 \cdot 5H_2O$$ is $$-x$$ kJ. The value of $$x$$ is ______ (nearest integer).

Using Hess's law:

$$CuSO_4(s) + aq \rightarrow CuSO_4(aq)$$, $$\Delta H_1 = -70$$ kJ/mol (dissolution of anhydrous)

$$CuSO_4 \cdot 5H_2O(s) + aq \rightarrow CuSO_4(aq)$$, $$\Delta H_2 = +12$$ kJ/mol (dissolution of hydrated)

We want: $$CuSO_4(s) + 5H_2O(l) \rightarrow CuSO_4 \cdot 5H_2O(s)$$, $$\Delta H_{hydration} = ?$$

By Hess's law: $$\Delta H_1 = \Delta H_{hydration} + \Delta H_2$$

$$-70 = \Delta H_{hydration} + 12$$

$$\Delta H_{hydration} = -70 - 12 = -82 \text{ kJ/mol}$$

So $$x = 82$$.

The answer is $$\boxed{82}$$.