Question

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 g mol$^{-1}$) in 600 g of water. Calculate the freezing point of the solution. (K$_f$ for water = 1.86 K kg mol$^{-1}$)

Hint:

For non-electrolytes such as glucose, urea and ethylene glycol, always take van't Hoff factor \(i=1\).

Answer 1

Concept: Freezing point depression is a colligative property and is given by: \[ \Delta T_f = iK_fm \] For ethylene glycol, which is a non-electrolyte: \[ i=1 \]

Step 1: Calculate moles of ethylene glycol. Mass given: \[ 31g \] Molar mass: \[ 62g\,mol^{-1} \] \[ \text{Moles} = \frac{31}{62} = 0.5 \]

Step 2: Calculate molality. Mass of water: \[ 600g = 0.600kg \] \[ m = \frac{0.5}{0.600} = 0.833 \] mol kg\(^{-1}\)

Step 3: Calculate depression in freezing point. \[ \Delta T_f = 1 \times 1.86 \times 0.833 \] \[ =1.55K \]

Step 4: Determine freezing point of solution. Freezing point of pure water: \[ 0^\circ C \] Hence \[ T_f = 0-1.55 \] \[ =-1.55^\circ C \]

Final Answer: \[ \boxed{-1.55^\circ C} \] The freezing point of the antifreeze solution is \[ \boxed{-1.55^\circ C} \]