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Pure Benzene, C6H6, freezes at 5.5°C and boils at 80.1°C. What is the boiling point of a solution consisting of cyclohexane(C6H12) dissolved in benzene if the freezing point of this solution is 0.0°C? [For benzene , Kf=5.12 °C/m , Kb = 2.53 °C/m ; For cyclohexane, Kf=20.0°C/m , Kb=2.79°C/m]
A.82.8°C
B.91.2°C
C.80.9°C
D.77.4°C
E.83.1°C

Sagot :

sebassandin

This problem is providing information about the freezing point depression of a solution consisting of benzene and cyclohexane and asks for the boiling point elevation for the same concentration of solute in the aforementioned solution, which turns out to be C. 80.9°C.

Colligative properties:

When a liquid homogeneous mixture or a solution is prepared by mixing a solute and and a solvent, the properties it has differ from those of the pure substances, which means we can find a freezing point depression, boiling point elevation and a couple more colligative properties, but these two are the subject of matter here.

Freezing point depression:

In this manner, we can start by using the freezing point depression in order to calculate the the molality of the solution, taking into account that freezing point depression constant depends on the solvent (cyclohexane) only:

[tex]T_f_{solution}-T_f_{solvent}=-m*Kf_{solvent}\\\\0.0\°C-5.5\°C=-m*20.0\°C/m\\\\m=\frac{-5.5\°C}{-20.0\°C/m} =0.275m[/tex]

By the way, in molal units.

Boiling point elevation:

Afterwards, we calculate the boiling point of the solution by using the formula of the boiling point elevation, which, again, depends on the boiling point elevation constant of the solvent:

[tex]T_b_{,\ solution}-T_b_{,\ solute}=m*Kb_{solvent}\\\\T_b_{,\ solution}-80.1\°C=0.275m*2.79\°C/m\\\\T_b_{,\ solution}=80.1\°C+0.275m*2.79\°C/m\\\\T_b_{,\ solution}=80.9\°C[/tex]

Learn more about colligative properties: https://brainly.com/question/4227527

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