Q86P - Chemistry: Molecular Nature Of Matter And Change

Question

Which aqueous solution has a freezing point closer to its predicted value, 0.01 m NaBr or 0.01 m MgCl₂? Explain.

Step-by-Step Solution

Answer

As compared to 0.01m ${\mathrm{MgCl}}_2,$ a 0.01m NaBr will have a value of a freezing point closer to its predicted value because of its lesser depression in freezing point.

Concept

Strong electrolytes in a solution dissociates into ions to give more number of particles. So, we include a multiplying factor in the equations for the colligative properties of electrolyte solutions, which is called van’t Hoff factor (i). 

Include the van't Hoff factor in the equation to determine the combinatorial features of strong electrolyte solutions: ${\mathbf{\Delta T}}_{\mathbf{f}}\mathbf{=}\mathbf{i}\left({\mathbf{K}}_{\mathbf{f}}\mathbf{m}\right).$

The freezing point depression is proportional to the molality and i. 

Calculate the van’t Hoff factor (i). for NaBr and MgCl 2 .

If strong electrolyte solutions behaved perfectly, the factor I would be equal to the product of the moles of particles in a solution and the moles of a dissolved solute, or 2 for NaBr and 3 for $MgC{l}_2$.

Depression in the freezing point for NaBr and M g C l 2 .

Using ${\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\left({\mathrm{K}}_{\mathrm{f}}\mathrm{m}\right)$

The molality is the same for both the solutions, so, the depression in the freezing point of the NaBr solution will be by 2 units, while in ${\mathrm{MgCl}}_2,$ the depression in the freezing point is by 3 units. As ${\mathrm{MgCl}}_2,$ has more particles, there will be more deviation from the ideal behavior.

So, NaBr will have a value of a freezing point closer to its predicted value because of its lesser depression in its freezing point.