Question

Calculate the boiling point elevation of an aqueous solution containing \(0.5\) g glucose in \(20\) g water if \(K_{b}\) for water is \(0.52\) \(\text{K kg mol}^{-1}\). [Molar mass of glucose \(=180\) \(\text{g mol}^{-1}\)]

• A. \(0.072K\)
• B. \(0.054K\)
• C. \(0.091K\)
• D. \(1.08K\)

Hint:

Find the molality of the glucose solution and multiply by Kb.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The elevation in boiling point is \(\Delta T_b=K_b\,m\), where \(m\) is the molality.

Step 2: Molality
Moles of glucose \(=\frac{0.5}{180}=2.78\times10^{-3}\).
Mass of water \(=20\) g \(=0.020\) kg.
\(m=\frac{2.78\times10^{-3}}{0.020}=0.139\ \text{mol kg}^{-1}\).

Step 3: Calculation
\(\Delta T_b=0.52\times0.139=0.072\) K, option (A).

Final Answer:
The boiling point elevation is about 0.072 K. \[ \boxed{\text{(A) }0.072K} \]