Question

How much ethyl alcohol must be added to 1 litre of water so that the solution will freeze at \(-14^\circ \text{C}\)? 

(K_f for water = 1.86 °C/mol)

• A. 7.5 mol
• B. 8.5 mol
• C. 9.5 mol
• D. 10.5 mol

Hint:

For freezing point depression problems: - Use the formula \( \Delta T_f = K_f \cdot m \). - Ensure the units of \( K_f \) and \( \Delta T_f \) are consistent. - Molality (\( m \)) is defined as moles of solute per kg of solvent.

Answer 1

The Correct Option is A

Step 1: Use the freezing point depression formula. The freezing point depression (\( \Delta T_f \)) is given by: \[ \Delta T_f = K_f \cdot m \] where: - \( \Delta T_f \) = freezing point depression, - \( K_f \) = cryoscopic constant (molal freezing point depression constant), - \( m \) = molality of the solution (moles of solute per kg of solvent). 
Step 2: Calculate the freezing point depression. The normal freezing point of water is \( 0^\circ {C} \), and the solution freezes at \(-14^\circ {C}\). Thus: \[ \Delta T_f = 0^\circ {C} - (-14^\circ {C}) = 14^\circ {C}. \] 
Step 3: Solve for molality (\( m \)). Using the formula: \[ \Delta T_f = K_f \cdot m, \] \[ 14 = 1.86 \cdot m. \] \[ m = \frac{14}{1.86} \approx 7.53 \, {mol/kg}. \] 
Step 4: Calculate the moles of ethyl alcohol required. Since the solvent is 1 litre of water, and the density of water is approximately \( 1 \, {kg/L} \), the mass of water is \( 1 \, {kg} \). Therefore: \[ {Moles of ethyl alcohol} = m \times {mass of solvent} = 7.53 \, {mol/kg} \times 1 \, {kg} = 7.53 \, {mol}. \] Conclusion: Approximately \( 7.5 \, {mol} \) of ethyl alcohol must be added to 1 litre of water to achieve the desired freezing point.