Question

The molal elevation constant for water is \(0.52\) \({}^{\circ }\)C kg mol\({}^{-1}\). Find out the boiling point of solution when \(6.0\) g of urea is dissolved in \(200\) g of water. (molar mass of urea \(=60\) g/mol)

• A. \(110.28\) \({}^{\circ }\)C
• B. \(100.26\) \({}^{\circ }\)C
• C. \(50.14\) \({}^{\circ }\)C
• D. \(0.26\) \({}^{\circ }\)C

Hint:

Find the molality of the urea solution, calculate the boiling point elevation, and add it to 100 degrees Celsius.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(\Delta T_b = K_b\, m\), and the boiling point is \(100 + \Delta T_b\) for aqueous solutions.

Step 2: Molality
Moles of urea \(= \dfrac{6.0}{60} = 0.1\) mol; water \(= 0.2\) kg.
\(m = \dfrac{0.1}{0.2} = 0.5\) mol kg\(^{-1}\).

Step 3: Calculation
\(\Delta T_b = 0.52\times 0.5 = 0.26\) \(^{\circ}\)C.
Boiling point \(= 100 + 0.26 = 100.26\) \(^{\circ}\)C.

Final Answer:
The solution boils at 100.26 degrees Celsius. \[ \boxed{\text{(B) }100.26^{\circ}\text{C}} \]