Question

Calculate the boiling point of an aqueous solution containing 18 g glucose in 100 g water if the molal elevation constant of water is \(0.5 \text{K kg mol}^{-1}\).
[Molar mass of glucose \(= 180 \text{g mol}^{-1}\) and boiling point of water \(= 100 ^{\circ}\text{C}\)]

• A. \(100 ^{\circ}\text{C}\)
• B. \(100.5 ^{\circ}\text{C}\)
• C. \(101.0 ^{\circ}\text{C}\)
• D. \(101.5 ^{\circ}\text{C}\)

Hint:

Delta Tb = Kb times molality; add to 100 C.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Boiling point elevation is \(\Delta T_b = K_b m\), where \(m\) is molality (mol per kg of solvent).

Step 2: Detailed Explanation:
Moles of glucose = \(18/180 = 0.1\) mol. Mass of water = 100 g = 0.1 kg.
\[ m = \frac{0.1}{0.1} = 1\ \text{mol/kg} \]
\[ \Delta T_b = 0.5 \times 1 = 0.5\ \text{K} \]
Boiling point of solution = \(100 + 0.5 = 100.5\ ^{\circ}\text{C}\).
Option (A) ignores the elevation. Options (C) and (D) would need a molality of 2 or 3.

Step 3: Final Answer:
\(100.5\ ^{\circ}\text{C}\), option (B).

Final Answer:
The boiling point becomes 100.5 C. \[ \boxed{\text{(B) }100.5\ ^{\circ}\text{C}} \]