KCET 2025 Chemistry Paper

To find the boiling point of the water solution, we need to calculate the boiling point elevation using the formula:$\Delta T_b = i \cdot K_b \cdot m$where:$\Delta T_b$ is the boiling point elevation, $i$ is the van 't Hoff factor, which is 1 for glucose as it does not dissociate in solution $K_b$ is the ebullioscopic constant for water, given as $0.52\ \mathrm{K\ kg\ mol}^{-1}$, $m$ is the molality of the solution.To calculate the molality $m$, use:$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$ For glucose $(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6)$, with a mass of 180 g :$\;\text{Moles of glucose}\; = \frac{180\ \mathrm{g}}{180\ \mathrm{g}/\mathrm{mol}} = 1\ \mathrm{mol}$The solvent (water) is 1 kg :$m = \frac{1\ \mathrm{mol}}{1\ \mathrm{kg}} = 1\ \mathrm{mol/kg}$ Substituting into the formula for $\Delta T_b$ :$\Delta T_b = 1 \times 0.52 \times 1 = 0.52\ \mathrm{K}$The normal boiling point of pure water is 373.15 K . Therefore, the boiling point of the solution is:$T_b = 373.15 + 0.52 = 373.67\ \mathrm{K}$