Given: (I) $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$$; $$\Delta H°_{298K} = -285.9$$ kJ mol$$^{-1}$$
(II) $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$$; $$\Delta H°_{298K} = -241.8$$ kJ mol$$^{-1}$$
The molar enthalpy of vapourisation of water will be :

We are given two reactions:

(I) $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$$ with $$\Delta H°_{298K} = -285.9$$ kJ mol$$^{-1}$$.

(II) $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$$ with $$\Delta H°_{298K} = -241.8$$ kJ mol$$^{-1}$$.

We need to find the molar enthalpy of vaporization of water, which is the enthalpy change for the reaction: $$H_2O(l) \rightarrow H_2O(g)$$.

Using Hess's law, we can combine the given reactions to get the desired reaction. Notice that reaction (I) produces liquid water, and reaction (II) produces water vapor. To go from liquid water to water vapor, we can reverse reaction (I) and then add reaction (II).

First, reverse reaction (I):

$$H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g)$$

When we reverse a reaction, the sign of $$\Delta H$$ changes. So, the enthalpy change for the reversed reaction is:

$$\Delta H_{\text{reverse I}} = -(-285.9) = +285.9$$ kJ mol$$^{-1}$$.

Now, add reaction (II) to this reversed reaction:

$$H_2O(l) \rightarrow H_2(g) + \frac{1}{2}O_2(g)$$ (from reversed I)

$$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$$ (from II)

Adding these two equations, the $$H_2(g)$$ and $$\frac{1}{2}O_2(g)$$ cancel out, leaving:

$$H_2O(l) \rightarrow H_2O(g)$$

According to Hess's law, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the steps:

$$\Delta H_{\text{vap}} = \Delta H_{\text{reverse I}} + \Delta H_{\text{II}} = 285.9 + (-241.8)$$

Now, perform the calculation:

$$285.9 - 241.8 = 44.1$$ kJ mol$$^{-1}$$.

Therefore, the molar enthalpy of vaporization of water is 44.1 kJ mol$$^{-1}$$.

Comparing with the options:

A. 241.8 kJ mol$$^{-1}$$

B. 22.0 kJ mol$$^{-1}$$

C. 44.1 kJ mol$$^{-1}$$

D. 527.7 kJ mol$$^{-1}$$

Hence, the correct answer is Option C.