Question

When 2.62 gof the non-volatile solid anthracene,${\mathbf{\text{C}}}_{\mathbf{14}}{\mathbf{\text{H}}}_{\mathbf{10}}$, is dissolved in 100.0 g of cyclohexane,${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{12}}$, the boiling point of the cyclohexane is raised by$\mathbf{0}\mathbf{.}{\mathbf{41}}^{\mathbf{0}}\mathit{C}$Calculate${\mathit{K}}_{\mathbf{\text{b}}}$ for cyclohexane.

Short answer

The value of $K_b$ for cyclohexane is $2.8^0{\text{Ckgmol}}^{-1}$.

Step-by-step solution

Given data.

Elevation in boiling point is$0.{41}^0\text{C}$.

The amount of anthracene is 2.62 g .

The concept of elevation of boiling point

When a solute is added to a solvent the boiling point of the solution increases. This is known as the elevation of boiling point. The boiling point of a solution containing a non-volatile solute is generally greater than the boiling point of the pure solvent.

Formula of elevation of boiling point

The formula of elevation of boiling point can be shown as:

$K_b=\frac{∆T_b}{m}$ (1)

where,

$∆T_b$is the elevation of boiling point, m is molality of the solution and ${\text{K}}_{\text{b}}$is the boiling point elevation constant.

Calculation of Kb (molal boiling point constant)

To find the value of molal boiling point constant, first we have to determine the number of moles of anthracene and the concentration of the solution in terms of molality.

$$\begin{gathered} \text{Moles of anthracene}=\frac{2.62\text{g}}{178.0{\text{gmol}}^{-1}} \\ =0.0147\text{mol} \end{gathered}$$

Molality can be determined by the formula as follows:

$$\begin{gathered} m=\frac{\text{number of moles of solute}}{\text{mass of the solvent in kilograms}} \\ m=\frac{0.0147\text{mol}}{100.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}} \\ m=0.147{\text{molkg}}^{-1} \end{gathered}$$

Puts the value of given data in equation (i).

$$\begin{gathered} K_b=\frac{0.{41}^0\text{C}}{0.147 {\text{molkg}}^{-1}} \\ =2.8^0 {\text{Ckgmol}}^{-1} \end{gathered}$$