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

Part (i): Calculation of freezing point

Concept: The depression in freezing point is a colligative property. It depends on the number of solute particles present in the solution. The formula for depression in freezing point is: \[ \boxed{\Delta T_f=iK_fm} \] where, \[ i=\text{Van't Hoff factor} \] \[ K_f=\text{molal depression constant} \] \[ m=\text{molality of solution} \]

Step 1: Calculate Van't Hoff factor Magnesium bromide undergoes complete dissociation: \[ MgBr_2 \rightarrow Mg^{2+}+2Br^- \] One molecule of MgBr$_2$ produces: \[ 1+2=3 \] ions. Therefore: \[ \boxed{i=3} \]

Step 2: Calculate moles of MgBr$_2$ Using: \[ \text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}} \] \[ =\frac{10.5}{184} \] \[ =0.0571\,mol \]

Step 3: Calculate molality Mass of water: \[ 250g=0.250kg \] Therefore: \[ m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}} \] \[ m=\frac{0.0571}{0.250} \] \[ m=0.2284\,mol\,kg^{-1} \]

Step 4: Calculate depression in freezing point \[ \Delta T_f=iK_fm \] Substituting the values: \[ \Delta T_f=3\times1.86\times0.2284 \] \[ \Delta T_f=1.27K \] The freezing point of pure water is: \[ 0^\circ C \] Hence: \[ T_f=0-1.27 \] \[ \boxed{T_f=-1.27^\circ C} \] Therefore, the freezing point of the solution is $-1.27^\circ C$.

Part (ii): Ideal and Non-ideal Solutions

Ideal solutions: An ideal solution is one which obeys Raoult's law over the entire range of concentration. For ideal solutions: \[ \Delta H_{mix}=0 \] and \[ \Delta V_{mix}=0 \] because intermolecular interactions between solute-solvent molecules are similar to solute-solute and solvent-solvent interactions.

Non-ideal solutions: Non-ideal solutions do not obey Raoult's law due to differences in intermolecular interactions. They show: \[ \Delta H_{mix}\neq0 \] and \[ \Delta V_{mix}\neq0 \] They may show positive or negative deviation from Raoult's law.

Final Answer:

(i) \[ \boxed{\text{Freezing point of MgBr}_2 \text{ solution}=-1.27^\circ C} \]

(ii) \[ \boxed{ \begin{array}{|c|c|} \hline \textbf{Ideal Solution} & \textbf{Non-ideal Solution}\\ \hline \text{Obeys Raoult's law} & \text{Does not obey Raoult's law}\\ \hline \Delta H_{mix}=0,\Delta V_{mix}=0 & \Delta H_{mix}\neq0,\Delta V_{mix}\neq0\\ \hline \end{array} } \]