Question

The boiling point elevation constant, kb, for CS2 is 2.3 o^C (degrees celsuis) and its boiling point is 46.3 o^C (degrees Celsius). What would be the boiling point of a one tenth saturated solution of S8 in CS2?
Thanks!

Answer 1

Answer on Question #80244 - Chemistry - General Chemistry

Question:

The boiling point elevation constant, $k_b$, for CS2 is 2.3 $O^{\wedge}C$ (degrees celsius) and its boiling point is 46.3 $O^{\wedge}C$ (degrees Celsius). What would be the boiling point of a one-tenth saturated solution of S8 in CS2?

Thanks!

Solution:

$$\Delta T_b = K_b \cdot b_8$$

- $\Delta T_b$, increase in the boiling point, is defined as $T_b$ (solution)-$T_b$ (pure solvent).

- $K_b$, the ebullioscopic constant, which depends on the properties of the solvent. It can be calculated as $K_b = RT_b2M / \Delta H_v$, where $R$ is the gas constant, and $T_b$ is the boiling point of the pure solvent [in K], $M$ is the molar mass of the solvent, and $\Delta H_v$ is the evaporation heat per mole of the solvent.

- $b_8$ is the molarity of the solution, calculated taking into account dissociation, since an increase in the boiling point is a colligative property, depending on the amount of particles in the solution.

$T_b$ (solution) = $T_b$ (pure solvent) + $K_b \cdot b_8 = 46.3 + 2.3 \cdot 1/10 = 46.3 + 0.23 = 46.53 \, \text{o}^{\wedge} \text{C}$ (degrees celsius).