The enthalpies of combustion of carbon and carbon monoxide are $$-393.5$$ and $$-283$$ kJ mol$$^{-1}$$ respectively. The enthalpy of formation of carbon monoxide per mole is

Step 1: Write the Thermochemical Equations

Based on the data provided, we can write out the chemical equations for the two combustion processes:

1. Combustion of Carbon (Graphite):

   $$\text{C}_{(s)} + \text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \quad \Delta H_1 = -393.5 \text{ kJ mol}^{-1}$$

2. Combustion of Carbon Monoxide:

   $$\text{CO}_{(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} \quad \Delta H_2 = -283 \text{ kJ mol}^{-1}$$

Step 2: Define the Target Equation

The standard enthalpy of formation of carbon monoxide ($$\text{CO}$$) corresponds to the reaction where $$1 \text{ mole}$$ of $$\text{CO}$$ is synthesized directly from its constituent elements in their standard states:

$$\text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)} \quad \Delta H_f^\circ = ?$$

Step 3: Apply Hess's Law of Constant Heat Summation

To match our target equation using the known reactions, we can subtract the second combustion equation from the first equation:

$$\text{Target Equation} = \text{Equation (1)} - \text{Equation (2)}$$

Mathematically executing this operation on their respective enthalpy changes:

$$\Delta H_f^\circ = \Delta H_1 - \Delta H_2$$

$$\Delta H_f^\circ = -393.5 \text{ kJ mol}^{-1} - (-283 \text{ kJ mol}^{-1})$$

$$\Delta H_f^\circ = -393.5 + 283 = -110.5 \text{ kJ mol}^{-1}$$

Conclusion:

The standard enthalpy of formation for one mole of carbon monoxide is determined to be $$-110.5 \text{ kJ}$$.

Answer: Option B — $$-110.5 \text{ kJ}$$