Question

An antifreeze solution is prepared by dissolving \(31g\) of ethylene glycol \((M=62g\,mol^{-1})\) in \(600g\) of water. Calculate the freezing point of the solution. \[ K_f\text{ for water}=1.86K\,kg\,mol^{-1} \]

Hint:

For depression in freezing point, use \(\Delta T_f=K_fm\), and final freezing point is \(0-\Delta T_f\) for water.

Answer 1

Concept:
Depression in freezing point is a colligative property. \[ \Delta T_f=K_fm \] where \(m\) is molality. Ethylene glycol is a non-electrolyte, so: \[ i=1 \]

Step 1: Calculate moles of ethylene glycol.
\[ \text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}} \] \[ \text{Moles}=\frac{31}{62} \] \[ \text{Moles}=0.5 \]

Step 2: Convert mass of water into kg.
\[ 600g=0.600kg \]

Step 3: Calculate molality.
\[ m=\frac{0.5}{0.600} \] \[ m=0.833 \]

Step 4: Calculate depression in freezing point.
\[ \Delta T_f=K_fm \] \[ \Delta T_f=1.86\times 0.833 \] \[ \Delta T_f=1.55K \]

Step 5: Calculate freezing point of solution.
Freezing point of pure water: \[ 0^\circ C \] So: \[ T_f=0-1.55 \] \[ T_f=-1.55^\circ C \] Hence: \[ \boxed{T_f=-1.55^\circ C} \]