Boiling Point Elevation Formula:

$ \Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b $

where:

$i_1$ and $i_2$ are the van't Hoff factors for ethylene glycol and glucose, respectively (both are 1 since they do not dissociate in solution).

$m_1$ and $m_2$ are the molalities of ethylene glycol and glucose, respectively.

$K_b$ is the ebullioscopic constant of water ($0.52 \, \text{K kg mol}^{-1}$).

Calculate Molality:

Each solute has 2 moles dissolved in 500 grams of water ($0.5 \, \text{kg}$):

$ m_1 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $

$ m_2 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $

Substitute into the Formula:

$ \Delta T_b = 1 \cdot 4 \cdot 0.52 + 1 \cdot 4 \cdot 0.52 = 4.16 $

Determine Boiling Point of Solution:

The normal boiling point of water is $373.16 \, \text{K}$.

Add the boiling point elevation to the normal boiling point:

$ T_b (\text{solution}) = 373.16 + 4.16 = 377.3 \, \text{K} $

Thus, the boiling point of the resulting solution is $377.3 \, \text{K}$.