Question

Boiling point of water at \(750\,\mathrm{mmHg}\) is \(99.63^\circ\mathrm{C}\). The amount of sucrose to be added to \(500\,\mathrm{g}\) water so that it boils at \(100^\circ\mathrm{C}\) is_ _ _ _ g. (Molar elevation constant \(K_b = 0.5\,\mathrm{K\,kg\,mol^{-1}}\))

Hint:

Boiling point elevation depends only on number of particles (colligative property), not on type of solute.

Answer 1

Correct Answer: 127

Step 1: Understanding the Concept:
Addition of a non-volatile solute like sucrose to a solvent increases its boiling point. This phenomenon is known as the elevation of boiling point (\(\Delta T_{b}\)), which is a colligative property.
Step 2: Key Formula or Approach:
Elevation of boiling point:
\[ \Delta T_{b} = K_{b} \times m = K_{b} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}} \]
Where:
\(w_{2}\) = mass of solute, \(M_{2}\) = molar mass of solute
\(w_{1}\) = mass of solvent in grams
: Detailed Explanation:
1. Calculate \(\Delta T_{b}\):
\[ \Delta T_{b} = T_{b}(\text{solution}) - T_{b}(\text{pure solvent}) \]
\[ \Delta T_{b} = 100^{\circ}\text{C} - 99.63^{\circ}\text{C} = 0.37^{\circ}\text{C} \text{ (or } 0.37\text{ K)} \]
2. Determine molar mass of sucrose (\(C_{12}H_{22}O_{11}\)):
\[ M_{2} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342\text{ g/mol} \]
3. Substitute values into the formula:
\[ 0.37 = 0.5 \times \frac{w_{2} \times 1000}{342 \times 500} \]
\[ 0.37 = 0.5 \times \frac{w_{2} \times 2}{342} = \frac{w_{2}}{342} \]
\[ w_{2} = 0.37 \times 342 \]
\[ w_{2} = 126.54\text{ g} \]
Rounding to the nearest integer, we get \(127\).
Step 3: Final Answer:
The amount of sucrose required is \(127\text{ g}\).