Chapter 16, Problem 76A

Interpretation Introduction

Interpretation: The boiling and the freezing point of $12\cdot 0\text{g}$ of naphthalene in $50.0\text{g}$ of benzene.

Concept Introduction:

When naphthalene (solute) is mixed with benzene (solvent), the boiling point (bp) and the freezing point (fp) of the solution are different from the bp and fp of the solute.

Under this condition the formula is:

  $\begin{array}{c}\Delta T=T_b-T_a\\ =K_{b\text{}}\text{}\times m\end{array}$   ...... (1)

  $\begin{array}{c}\Delta T=T_f-T_c\\ =K_{b\text{}}\text{}\times m\end{array}$   ...... (2)

Here $T_f$ is the freezing point, $T_a$ is the normal boiling point, $T_c$ is the normal freezing point, $K_f$ is the molal freezing point depression constant, $m$ is molality, $T_b$ is the boiling point, $K_b$ is the molal boiling point elevation constant.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (3)

The expression for molality is as follows:

  $m=\frac{\text{moles of solute}}{\text{Mass of solvent(kg)}}$   ...... (4)

Answer to Problem 76A

The change in boiling point is $85.08°\text{C}$ and the change in freezing point of $-4.13°\text{C}$ for a solution containing $12.0\text{g}$ in $50.0\text{g}$ of benzene.

Explanation of Solution

Mass of naphthalene is $12.0\text{g}$ .

Mass of benzene is $50.0\text{g}$ .

The boiling point constant of benzene is $2.65°\text{C/m}$

The freezing point depression of benzene is $-5.12°\text{C/m}$ .

The normal boiling point is $80.1°\text{C}$

The normal freezing point $-5.5°\text{C}$ .

Substitute $12\cdot 0\text{g}$ for mass of naphthalene $\left({\text{C}}_{10}{\text{H}}_8\right)$ and $128.17\text{g/mol}$ for molar mass of $\left({\text{C}}_{10}{\text{H}}_8\right)$ in equation (3).

  $\begin{array}{c}\text{moles}\left({\text{C}}_{10}{\text{H}}_8\right)=\frac{\text{ 12 g}}{\text{128}\text{.17 g/mol}}\\ =0.094\text{ mol}\end{array}$

Substitute $0.094\text{mol}$ for solute and $0.05\text{kg}$ for mass of solvent in equation (4).

  $\begin{array}{c}m=\frac{0.094}{0.05}\\ =0.8m\end{array}$

The molal boiling point elevation constant of water is $2.65°{\text{Cmol}}^{-1}$ .

The molal freezing point depression constant of water is $-5.12°{\text{Cmol}}^{-1}$ .

Substitute $2.65°{\text{Cmol}}^{-1}$ for $K_b$ , $80.1°\text{C}$ for $T_c$ , and $1.88$ for $m$ in equation (2).

  $\begin{array}{c}{T}_b-80.1°\text{C}=2.65°\text{C/m}\times 1.88\\ T_b=85.08°\text{C}\end{array}$

Substitute $-5.12°{\text{Cmol}}^{-1}$ for $K_a$ , $-5.5°\text{C}$ for $T_a$ , and $1.88$ for $m$ in equation (1).

  $\begin{array}{c}{T}_f-\left(-5.5°\text{C}\right)=-5.12°{\text{Cmol}}^{-1}\times 1.88\\ T_f=-4.13°\text{C}\end{array}$

Conclusion

The change in boiling point is $85.08°\text{C}$ and the change in freezing point of $-4.13°\text{C}$ for a solution containing $12.0\text{g}$ in $50.0\text{g}$ of benzene.