Question

Pure toluene $\left({\text{C}}_7{\text{H}}_8\right)$has a normal boiling point of$\mathbf{110}\mathbf{.}{\mathbf{60}}^{\mathbf{0}}\mathbf{\text{C}}$. A solution of 7.80 anthracene$\left({\text{C}}_{14}{\text{H}}_{10}\right)$in 100.0 gof toluene has a boiling point of$\mathbf{112}\mathbf{.}{\mathbf{06}}^{\mathbf{0}}\mathbf{\text{C.}}$Calculate${\mathit{K}}_{\mathbf{\text{b}}}$for toluene.

Short answer

The value of $K_b$ is $3.33{\text{Kkgmol}}^{-1}$.

Step-by-step solution

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${\text{K}}_{\text{b}}$ ia the boiling point elevation constant.

Determination of elevation of boiling point

The elevation of the boiling point can be determined by the formula:

$∆T=T_2-T_1$

Substituting the value of $T_1$and $T_2$in the above equation, we get:

$$\begin{gathered} ∆T_b=112.{06}^0\text{C}-110.{60}^0\text{C} \\ =(112.06+273)\text{K}-(110.60+273)\text{K} \\ =1.46\text{K} \end{gathered}$$

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.

$$\begin{gathered} \text{Moles of anthracene}=\frac{7.80\text{g}}{178.0{\text{gmol}}^{-1}} \\ =0.0438\text{mol} \end{gathered}$$

Molality can be determined by the formula as follows:

$$\begin{gathered} m=\frac{\text{Number} \text{of} \text{moles of solute}}{\text{mass of solvent in kilogram}} \\ m=\frac{0.0438\text{mol}}{100.0\text{g×}\frac{1\text{kg}}{1000\text{g}}} \\ m=0.438{\text{molkg}}^{-1} \end{gathered}$$

Put the value of given data in equation (1).

$$\begin{gathered} K_b=\frac{1.46\text{K}}{0.438 {\text{molkg}}^{-1}} \\ =3.33 {\text{Kkgmol}}^{-1} \end{gathered}$$