Q93P - Chemistry: Molecular Nature Of Matter And Change

Question

Which solution has the higher boiling point?

(a) 38.0 g of ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}$  in 250. g of ethanol or 38.0 g of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$ in 250. g of ethanol

(b) 15 g of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$  in 0.50 kg of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ or 15 g of NaCl in 0.50 kg of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

Step-by-Step Solution

Answer

1. A $\mathbf{38}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{250}\mathbf{ }\mathbf{g}$ of ethanol will have a higher boiling point than a  $\mathbf{38}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{250}\mathbf{ }\mathbf{g}$ of ethanol.
2. A  $\mathbf{15}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{NaCl}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{kg}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ will have a higher boiling point than a $\mathbf{15}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{kg}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

Elevation in boiling point

The magnitude of a boiling-point elevation is directly proportional to the molality or the molal concentration of a solute as ${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\mathrm{m}$

(a) Calculate the molality of both the given solutions

$\mathrm{molality} = \frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{kg} \mathrm{of} \mathrm{solvent} \left(\mathrm{mass} \mathrm{of} \mathrm{solvent}\right)}$

$$\begin{gathered} \mathrm{Molality} \mathrm{of} {\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3=\left[\frac{38 \mathrm{g} {\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3}{92 \mathrm{g} {\mathrm{C}}_3{\mathrm{H}}_8{\mathrm{O}}_3}\right]\left[\frac{1000 \mathrm{g}}{250 \mathrm{g} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}\right] \\ =1.652 \mathrm{m}. \end{gathered}$$

$$\begin{gathered} \mathrm{Molality} \mathrm{of} {\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2=\left[\frac{38 \mathrm{g} {\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2}{62 \mathrm{g} {\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2}\right]\left[\frac{1000 \mathrm{g}}{250 \mathrm{g} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}}\right] \\ =2.451 \mathrm{m}. \end{gathered}$$

Boiling-point elevation in both the solutions

A boiling-point elevation is directly proportional to the molality of a solute. The molality of the  $C_3{H}_8{O}_3$ as calculated above is 1.652 m, while that of the  $C_2{H}_6{O}_2$ is 2.451 m.

The molality of  $C_2{H}_6{O}_2$ is more, so the elevation in the boiling point will be more, and hence  $C_2{H}_6{O}_2$ will have a higher boiling point.