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

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

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

Step 1: Calculate moles of solute.
The number of moles of MgBr\(_2\) is calculated by dividing the given mass by the molar mass: \[ \text{Moles of MgBr}_2 = \frac{10.5}{184} = 0.0571 \, \text{mol} \]
Step 2: Calculate molality.
The mass of water is given as 250 g, which is equivalent to 0.25 kg. The molality (\(m\)) is calculated by dividing the moles of solute by the mass of the solvent in kilograms: \[ m = \frac{0.0571}{0.25} = 0.2284 \, \text{mol kg}^{-1} \]
Step 3: Determine the Van’t Hoff factor.
MgBr\(_2\) dissociates into 3 ions (1 Mg\(^{2+}\) and 2 Br\(^-\)): \[ MgBr_2 \rightarrow Mg^{2+} + 2Br^- \] Thus, the Van't Hoff factor (\(i\)) is 3: \[ i = 3 \]
Step 4: Calculate depression in freezing point.
Now, use the formula for depression in freezing point: \[ \Delta T_f = 3 \times 1.86 \times 0.2284 = 1.27 \, K \]
Step 5: Calculate the freezing point of the solution.
The freezing point of water is 0°C, so the freezing point of the solution is: \[ T_f = 0 - 1.27 = -1.27^\circ C \]
Final Answer:
\[ \boxed{-1.27^\circ C} \]