Question

Calculate the freezing point of a solution when \(10.5\) g of MgBr\(_2\) was dissolved in 250 g of water, assuming MgBr\(_2\) undergoes complete dissociation. (Given: Molar mass of MgBr\(_2\) = 184 g mol\(^{-1}\), \(K_f\) for water = 1.86 K kg mol\(^{-1}\))

Hint:

Electrolytes increase depression due to higher Van’t Hoff factor.

Answer 1

Concept: Depression in freezing point is given by: \[ \Delta T_f = i K_f m \] Where:

• \(i\) = Van’t Hoff factor
• \(m\) = molality

Step 1: Calculate moles of solute. \[ \text{Moles of MgBr}_2 = \frac{10.5}{184} = 0.0571 \, \text{mol} \]
Step 2: Calculate molality. Mass of water = 250 g = 0.25 kg \[ m = \frac{0.0571}{0.25} = 0.2284 \, \text{mol kg}^{-1} \]
Step 3: Van’t Hoff factor. MgBr\(_2\) dissociates as: \[ MgBr_2 \rightarrow Mg^{2+} + 2Br^- \] Total ions = 3 \[ i = 3 \]
Step 4: Calculate depression in freezing point. \[ \Delta T_f = 3 \times 1.86 \times 0.2284 \] \[ \Delta T_f = 1.27 \, K \]
Step 5: Freezing point of solution. \[ T_f = 0 - 1.27 = -1.27^\circ C \]
Final Answer: \[ \boxed{-1.27^\circ C} \]