Question

The freezing point depression of a solution containing \(0.6\ \text{g}\) of urea \((\text{molar mass}=60\ \text{g mol}^{-1})\) in \(100\ \text{g}\) of benzene is in K \((K_f=4.0\ \text{K kg mol}^{-1})\):

• A. \(0.30\)
• B. \(0.58\)
• C. \(0.40\)
• D. \(0.24\)

Hint:

For non-electrolytes like urea, \[ \Delta T_f=K_fm \] where molality is calculated using mass of solvent in kg, not total solution mass.

Answer 1

The Correct Option is C

Step 1: List the given data.
Mass of urea = 0.6 g
Molar mass of urea (NH2CONH2) = 60 g/mol
Mass of benzene (solvent) = 100 g = 0.1 kg
Cryoscopic constant of benzene: \(K_f = 4.0\) K kg/mol

Step 2: Calculate the number of moles of urea.
\[ n_{\text{urea}} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.6\ \text{g}}{60\ \text{g/mol}} = 0.01\ \text{mol} \]

Step 3: Calculate the molality of the solution.
Molality is defined as moles of solute per kilogram of solvent: \[ m = \frac{n_{\text{solute}}}{\text{mass of solvent (kg)}} = \frac{0.01\ \text{mol}}{0.1\ \text{kg}} = 0.1\ \text{mol/kg} \]

Step 4: Apply the freezing point depression formula.
The freezing point depression is given by: \[ \Delta T_f = K_f \times m \]
This colligative property occurs because the solute particles disrupt the ordered crystal lattice of the freezing solvent, requiring a lower temperature for solidification to occur.

Step 5: Calculate the freezing point depression.
\[ \Delta T_f = 4.0\ \text{K kg/mol} \times 0.1\ \text{mol/kg} = 0.40\ \text{K} \]

Step 6: Interpret and state the final answer.
The freezing point of benzene (normally 278.65 K or 5.5 degrees C) is depressed by 0.40 K when 0.6 g of urea is dissolved in 100 g of benzene. Urea is a non-electrolyte (van't Hoff factor i = 1), so no correction factor is needed.
\[ \boxed{\Delta T_f = 0.40\ \text{K}} \]