Question

What is the boiling point of the solution containing $2.33g$ of caffeine ${C_4}{H_{10}}{N_4}{O_2}$ , dissolved in $15.0g$ of benzene? The boiling point of pure benzene is ${80.01^ \circ }C$ and boiling point elevation constant, ${K_b}$ is ${2.53^ \circ }C{m^{ - 1}}$ .

Answer 1

The solute-to-solvent ratio, but not the identity of the solute, determines boiling point elevation, which is a colligative property of matter. This means that the amount of solute added to a solution affects the boiling point of the solution. The higher the solute concentration in a solution, the higher the boiling point elevation.
$\Delta {T_b} = {K_b}m$
$\Delta {T_b}$ = is the boiling point elevation
${K_b}$ = is the boiling point elevation constant
$m$ = is the morality of the solution.

Complete answer:
Given:
Mass ${C_4}{H_{10}}{N_4}{O_2} = 2.33g$
Mass of benzene $15.0g = 0.015kg$
To find: Boiling point of solution
Molar mass of caffeine ${C_4}{H_{10}}{N_4}{O_2}$ $= 194.19{\text{ }}gmo{l^{ - 1}}$
First, we need to calculate number of moles of ${C_4}{H_{10}}{N_4}{O_2}$ ,
Number of moles of ${C_4}{H_{10}}{N_4}{O_2}$ $= \dfrac{{Given{\text{ Mass}}}}{{Molar{\text{ Mass}}}}$
Number of moles of ${C_4}{H_{10}}{N_4}{O_2}$ $= \dfrac{{2.33g}}{{194.19gmo{l^{ - 1}}}}$
Number of moles of ${C_4}{H_{10}}{N_4}{O_2}$ $= 0.012mol$
Molality $= \dfrac{{moles{\text{ of solute}}}}{{kg{\text{ of solvent}}}}$
Molality $= \dfrac{{0.012\;mol}}{{0.015\;kg}}$
Molality $= 0.8$
Now, we calculate the boiling point elevation by using the following formula,
$\Delta {T_b} = {K_b}m$
$\Delta {T_b} = 2.53 \times 0.8$
$\Delta {T_b} = {2.02^ \circ }C$
Finally, we calculate the boiling point,
${T_b} = T_b^\circ + \Delta {T_b}$
${T_b} = {80.1^ \circ }C + {2.02^ \circ }C$
${T_b} \Rightarrow 82.1{\text{ }}^\circ C$ .

Additional Information:
The temperature at which a liquid's vapour pressure equals the pressure of its surroundings is known as the boiling point. Non-volatile compounds do not evaporate easily and have extremely low vapour pressures (assumed to be zero). When a non-volatile solute is added to a solvent, the resulting solution has a lower vapour pressure than the pure solvent.

Note:
The pressure of a liquid's surroundings also affects its boiling point (which is why water boils at temperatures lower than ${100^ \circ }C$ at high altitudes, where the surrounding pressure is low).