Question

Calculate the boiling point of solution prepared by adding \(2.93\) g of solute to \(0.059\) kg of solvent, if the boiling point of pure solvent = \(61.2^{\circ }C\), \(K_{b}=3.85^{\circ }C\) Kg mol\({}^{-1}\) and molar mass of solute \(=32\) g/mol)

• A. \(64.13{}^{0}C\)
• B. \(58.27{}^{0}C\)
• C. \(67.17{}^{0}C\)
• D. \(63.45{}^{0}C\)

Hint:

Find the molality, multiply by Kb to get the elevation, and add it to the boiling point of the pure solvent.

Answer 1

The Correct Option is C

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 solute \(=\dfrac{2.93}{32}=0.0916\,\text{mol}\).
\(m=\dfrac{0.0916}{0.059}=1.552\,\text{mol kg}^{-1}\)

Step 3: Elevation
\(\Delta T_b=3.85\times1.552=5.97^{\circ}\text{C}\)

Step 4: Boiling Point
\(T_b=61.2+5.97=67.17^{\circ}\text{C}\), option (C).

Final Answer:
The boiling point of the solution is \(67.17^{\circ}\text{C}\). \[ \boxed{\text{(C) }67.17{}^{0}C} \]