Question

A scientist wants to perform an experiment in aqueous solution in a hill station where the boiling point of water is 98.98°C. How much urea $(mol.wt \,60 g\, mol^{-1})$ is to be added by him to 2 kg of water to get the boiling point 100°C at the same place? ($K_b$ of water=0.51$K kg mol^{-1}$)

• A. 240 g
• B. 120 g
• C. 180 g
• D. 60 g
• E. 1.02 g

Answer 1

The Correct Option is A

Boiling Point Elevation 

Given:

• Boiling point of water at hill station = 98.98°C
• Desired boiling point = 100°C
• Elevation in boiling point, \( \Delta T_b = 100 - 98.98 = 1.02\, K \)
• Molal elevation constant, \( K_b = 0.51\, K \cdot kg \cdot mol^{-1} \)
• Mass of water = 2 kg
• Molar mass of urea = 60 g/mol

Using the formula:

\[ \Delta T_b = K_b \cdot m \]

Where \( m \) is molality = moles of solute / kg of solvent

So,

\[ m = \frac{\Delta T_b}{K_b} = \frac{1.02}{0.51} = 2 \, mol/kg \]

Since water is 2 kg:

\[ \text{Moles of urea} = 2 \times 2 = 4 \, mol \]

\[ \text{Mass of urea} = 4 \times 60 = \mathbf{240 \, g} \]

Correct Answer: 240 g