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: \[ \text{Molar mass of MgBr}_2 = 184 \, g\,mol^{-1} \] \[ K_f \text{ for water} = 1.86 \, K\,kg\,mol^{-1} \] (ii) Write two differences between ideal and non-ideal solutions.

Hint:

For electrolytes: \[ \boxed{\Delta T_f=iK_fm} \] Always calculate the Van't Hoff factor first. Examples: \[ NaCl \rightarrow Na^+ + Cl^- \Rightarrow i=2 \] \[ MgBr_2 \rightarrow Mg^{2+}+2Br^- \Rightarrow i=3 \] \[ \boxed{\text{More ions } \Rightarrow \text{greater depression in freezing point}} \]

Answer 1

Step 1: Van't Hoff factor and moles.
$MgBr_2 \rightarrow Mg^{2+} + 2Br^-$ gives $i = 3$. Moles of $MgBr_2 = \frac{10.5}{184} = 0.0571\,mol$. Molality $m = \frac{0.0571}{0.250\,kg} = 0.228\,mol\,kg^{-1}$.
Step 2: Freezing point depression.
\[ \Delta T_f = i \cdot K_f \cdot m = 3 \times 1.86 \times 0.228 = 1.27\,K \] Freezing point of solution $= 0 - 1.27 = -1.27^\circ C$.
Step 3: Ideal vs non-ideal solutions (two differences).
Ideal solutions obey Raoult's law over all concentrations; $\Delta H_{mix} = 0$ and $\Delta V_{mix} = 0$ because solute-solvent interactions equal pure-component interactions. Non-ideal solutions disobey Raoult's law; $\Delta H_{mix} \neq 0$ and $\Delta V_{mix} \neq 0$ due to unequal intermolecular forces, showing positive or negative deviations.
\[ \boxed{T_f = -1.27^\circ C} \]