2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

<p><strong>Boiling Point Elevation Formula</strong>:</p> <p>$ \Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b $</p> <p>where:</p></p> <p><p>$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).</p></p> <p><p>$m_1$ and $m_2$ are the molalities of ethylene glycol and glucose, respectively.</p></p> <p><p>$K_b$ is the ebullioscopic constant of water ($0.52 \, \text{K kg mol}^{-1}$).</p></p> <p><strong>Calculate Molality</strong>:</p> <p><p>Each solute has 2 moles dissolved in 500 grams of water ($0.5 \, \text{kg}$):</p> <p>$ m_1 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $</p> <p>$ m_2 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $</p></p> <p><p><strong>Substitute into the Formula</strong>:</p> <p>$ \Delta T_b = 1 \cdot 4 \cdot 0.52 + 1 \cdot 4 \cdot 0.52 = 4.16 $</p></p> <p><p><strong>Determine Boiling Point of Solution</strong>:</p></p> <p><p>The normal boiling point of water is $373.16 \, \text{K}$.</p></p> <p><p>Add the boiling point elevation to the normal boiling point:</p> <p>$ T_b (\text{solution}) = 373.16 + 4.16 = 377.3 \, \text{K} $</p></p> <p>Thus, the boiling point of the resulting solution is $377.3 \, \text{K}$.</p>