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

Hint: 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).