Question

A solution of urea (mol mass $56\, g\, mol^{-1})$ boils at $100.18 ^\circ\,C$ at the atmospheric pressure. If $K_f$ and $K_b$ for water are $1.86$ and $0.512\, K\, kg$ $mol^{-1}$ the above solution will freeze at

• A. -6.54$^\circ$C
• B. 6.54$^\circ$C
• C. 0.654$^\circ$C
• D. -0.654$^\circ$C

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the freezing point of the urea solution using the depression in the freezing point concept. Given data includes:

• Molar mass of urea: 56\, g\, mol^{-1}
• Boiling point of the solution: 100.18 ^\circ\,C
• Boiling point elevation constant: K_b = 0.512\, K\, kg\, mol^{-1}
• Freezing point depression constant: K_f = 1.86\, K\, kg\, mol^{-1}

First, we will calculate the molality of the solution using the elevation in boiling point formula:

The boiling point elevation formula is given by:

\Delta T_b = K_b \cdot m \cdot i

where \Delta T_b is the change in boiling point, m is the molality of the solution, and i is the van't Hoff factor (which is 1 for non-electrolytes like urea).

The change in boiling point, \Delta T_b, is:

\Delta T_b = 100.18 - 100.0 = 0.18\, ^\circ C

Now, substitute the values into the boiling point elevation formula:

0.18 = 0.512 \cdot m

Solving for m:

m = \frac{0.18}{0.512} \approx 0.3516\, molal

Next, we use this molality to find the freezing point depression:

The freezing point depression formula is:

\Delta T_f = K_f \cdot m \cdot i

Substituting the known values:

\Delta T_f = 1.86 \cdot 0.3516 \approx 0.654\, ^\circ C

This calculated depression in freezing point means that the freezing point of the solution is:

0 - 0.654 = -0.654\, ^\circ C

Therefore, the solution will freeze at -0.654 ^\circ C, which corresponds to the correct option:

• -0.654^\circC