MP Board · Class 12 · Chemistry · SolutionsCalculate the boiling point elevation for a solution prepared by dissolving 18 g of glucose ($C6H{12}O6$) in 1 kg of water. ($Kb$ for water = $0.52 \text{ K kg mol}^{-1}$).

Given Data:

• Mass of solute (glucose, $W_2$) = $18 \text{ g}$
• Mass of solvent (water, $W_1$) = $1 \text{ kg} = 1000 \text{ g}$
• Molar mass of glucose ($C_6H_{12}O_6$, $M_2$) = $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \text{ g mol}^{-1}$
• Ebullioscopic constant for water ($K_b$) = $0.52 \text{ K kg mol}^{-1}$

Formula:\nElevation in boiling point ($\Delta T_b$) is given by the formula:

$$\Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1}$$

Step-by-Step Calculation:

1. Substitute the given values into the formula: $$\Delta T_b = \frac{0.52 \times 18 \times 1000}{180 \times 1000}$|
2. Simplify the terms: $$\Delta T_b = \frac{0.52 \times 18}{180}$$
3. Cancel out common factors ($18 / 180 = 1 / 10$): $$\Delta T_b = \frac{0.52}{10}$|
4. Calculate the final value: $$\Delta T_b = 0.052 \text{ K}$$ (or °C)

Answer:\nThe elevation in the boiling point of the given solution is $0.052 \text{ K}$.