Chapter 13, Problem 67QP

Interpretation Introduction

Interpretation:

The normal freezing point and boiling point of the given solution are to be calculated with the help of mass of solute and volume of solvent.

Concept introduction:

The expression for freezing point depression is given as follows:

$T_{\text{f}}^{°}-T_{\text{f}}=i{K}_{\text{f}}m$

Here, $\Delta T_{\text{f}}$

denotes freezing point depression of solute, $i$

denotes Van’t Hoff factor, $K_{\text{f}}$

denotes the freezing point depression constant $\left(1.99°\text{C/}m\right)$, and $m$ is the molality of solute.

The freezing point of pure solute is represented by $T_{\text{f}}^{°}$, and the freezing point of solution is represented by $T_{\text{f}}$.

The expression for boiling point elevation is given as follows:

$T_{\text{b}}-T_{\text{b}}^{°}=i{K}_{\text{b}}m$

Here, the boiling point of pure solute is represented by $T_{\text{b}}^{°}$, the boiling point of solution is represented by $T_{\text{b}}$, $i$

denotes Van’t Hoff factor, $K_{\text{b}}$

denotes the boiling point elevation constant $\left(0.52°\text{C/}m\right)$, and $m$ is the molality of solute.

$\text{1000 mL}$ is equivalent to 1 kg

The molality of the solution is calculated by the expression as:

$m=\frac{\text{moles of solute}}{\text{mass of solvent }\left(\text{kg}\right)}$

Moles of compound can be calculated by the expression:

$\text{moles = }\frac{\text{mass of compound}}{\text{molar mass of compound}}$

So, molality can also be calculated by:

$m=\left(\frac{\text{mass of solute}}{\text{molar mass of}\text{solute}}\right)\left(\frac{\text{1}}{\text{mass of solvent }\left(\text{kg}\right)}\right)$

Answer to Problem 67QP

Solution:

$\begin{array}{l}{T}_{\text{f}}=-10.0°\text{C}\\ T_{\text{b}}=102.8°\text{C}\end{array}$

$\begin{array}{l}{T}_{\text{f}}=-7.14°\text{C}\\ T_{\text{b}}=102.0°\text{C}\end{array}$

Explanation of Solution

a) $\text{21}\text{.2 g NaCl in 135 mL of water}$

Since $\text{1000 mL}$ is equivalent to 1 kg, the mass of $\text{135 mL}$ of water is $\text{0}\text{.135 kg}$.

For $\text{NaCl}$, the value of $i=2$.

The molar mass of $\text{NaCl}$ is $58.4\text{ g/mol}$.

The normal boiling point and freezing point of water are given as follows:

$\begin{array}{l}{T}_{\text{f}}^{°}=0°\text{C}\\ T_{\text{b}}^{°}=100°\text{C}\end{array}$

The molality of the solution is calculated by the expression as follows:

$\begin{array}{c}m=\frac{\text{moles of NaCl}}{\text{mass of water }\left(\text{kg}\right)}\\ =\left(\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}}\right)\left(\frac{\text{1}}{\text{mass of water }\left(\text{kg}\right)}\right)\end{array}$

Substitute the required values in the equation given above,

$\begin{array}{c}m=\left(\frac{\text{21}\text{.2 g}}{\text{58}\text{.4 g/mol}}\right)\left(\frac{\text{1}}{\text{0}\text{.135 kg}}\right)\\ =2.70\text{ m}\end{array}$

Therefore, the molality of $\text{NaCl}$

is $2.70\text{ m}$.

The expression for freezing point depression is given as follows:

$T_{\text{f}}^{°}-T_{\text{f}}=i{K}_{\text{f}}m$

Substitute the values of $T_{\text{f}}^{°},i,K_{\text{f}}$, and $m$ in the equation given above.

$\begin{array}{c}0°\text{C}-T_{\text{f}}=2\times \left(1.86°\text{C/}m\right)\times 2.69m\\ T_{\text{f}}=-10.0°\text{C}\end{array}$

Hence, the normal freezing point of $\text{NaCl}$

is $-10.0°\text{C}$.

The expression for boiling point elevation is given as follows:

$T_{\text{b}}-T_{\text{b}}^{°}=i{K}_{\text{b}}m$

Substitute the values of $T_{\text{b}}^{°},i,K_{\text{b}}$, and $m$ in the equation given above,

$\begin{array}{c}{T}_{\text{b}}-100°\text{C}=2\times \left(0.52°\text{C/}m\right)\times \left(2.69m\right)\\ T_{\text{b}}-100°\text{C}=2.80°\text{C}\\ T_{\text{b}}=102.8°\text{C}\end{array}$

Hence, the normal boiling point of $\text{NaCl}$

is $102.8°\text{C}$.

b) $15.4\text{ g urea in }66.7\text{ mL of water}$

Since $\text{1000 mL}$ is equivalent to 1 kg, the mass of $\text{66}\text{.7 mL}$

of water is $\text{0}\text{.0667 kg}$.

For urea, the value of $i=1$.

The molar mass of urea is $60.06\text{ g/mol}$.

The normal boiling point and freezing point of water are given as follows:

$\begin{array}{l}{T}_{\text{f}}^{°}=0°\text{C}\\ T_{\text{b}}^{°}=100°\text{C}\end{array}$

The molality of the solution can be evaluated as follows:

$\begin{array}{c}m=\frac{\text{moles of urea}}{\text{mass of water }\left(\text{kg}\right)}\\ =\left(\frac{\text{mass of urea}}{\text{molar mass of urea}}\right)\left(\frac{\text{1}}{\text{mass of water }\left(\text{kg}\right)}\right)\end{array}$

Substitute the required values in the equation given above.

$\begin{array}{c}m=\left(\frac{\text{15}\text{.4 g}}{\text{60}\text{.06 g/mol}}\right)\left(\frac{\text{1}}{\text{0}\text{.0667 kg}}\right)\\ =3.84\text{ m}\end{array}$

Therefore, molality of urea is $3.84\text{ m}$.

The expression for freezing point depression is given as follows:

$T_{\text{f}}^{°}-T_{\text{f}}=i{K}_{\text{f}}m$

Substitute the values of $T_{\text{f}}^{°},i,K_{\text{f}}$, and $m$ in the equation given above,

$\begin{array}{c}0°\text{C}-T_{\text{f}}=1\times \left(1.86°\text{C/}m\right)\times 3.84m\\ T_{\text{f}}=-7.14°\text{C}\end{array}$

Hence, the normal freezing point of urea is $-7.14°\text{C}$.

The expression for boiling point elevation is given as follows:

$T_{\text{b}}-T_{\text{b}}^{°}=i{K}_{\text{b}}m$

Substitute the values of $T_{\text{b}}^{°},i,K_{\text{b}}$, and $m$ in the equation given above,

$\begin{array}{c}{T}_{\text{b}}-100°\text{C}=1\times \left(0.52°\text{C/m}\right)\times \left(\text{3}\text{.84 m}\right)\\ T_{\text{b}}-100°\text{C}=1.99°\text{C}\\ T_{\text{b}}=102.0°\text{C}\end{array}$

Hence, the normal boiling point of urea is $102.0°\text{C}$.