Chapter 13, Problem 65QP

Interpretation Introduction

Interpretation:

The freezing point of the solution, formed by condensing given volume of a gas into $75.0\text{ g}$ of benzene, is to be calculated.

Concept introduction:

The ideal gas equation is given as follows:

$PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the absolute temperature.

The temperature at which change of liquid state to solid state occurs is called freezing point.

The freezing point depression is as follows:

$\Delta T_{\text{f}}=m\times K_{\text{f}}$

Here, $\Delta T_{\text{f}}$ is the change in temperature of the freezing point in $\text{C}$, $m$ is the molality of the solution, and $K_{\text{f}}$ is themolal freezing point depression constant for the solvent.

For benzene

$K_{\text{f}}=5.12\frac{\text{C}}{m}$

The actual freezing point is the difference between freezing point of pure solvent and freezing point depression.

$T_f=T_f{}^{°}-\Delta T_{\text{f}}$

Convert $\text{mm Hg}$ to $\text{atm}$, since

$1\text{ mm Hg}=\frac{1}{760}\text{ atm}$

Thus, conversion factor will be $\left(\frac{\frac{1}{760}\text{ atm}}{\text{ 1 mm Hg}}\right)$.

Convert $\text{g}$ to kilograms $\left(\text{kg}\right)$, since, $\text{1 g}={10}^{-3}\text{ kg}$

Thus, the correct conversion factor will be $\left(\frac{{10}^{-3}\text{ kg}}{\text{1 g}}\right)$.

Answer to Problem 65QP

Solution: $-5.4{}^{°}\text{C}$.

Explanation of Solution

Given information: Volume of a gas is $4.00\text{ L}$.

Pressure is $748\text{ mm Hg}$.

Absolute Temperature is ${\text{27 }}^{°}\text{C}$.

Mass of benzene is $75.0\text{ g}$.

The conversion factor for $\text{mm Hg}$ to $\text{atm}$ is $\left(\frac{\frac{1}{760}\text{ atm}}{\text{ 1 mm Hg}}\right)$.

Multiply the conversion factor with the given quantity, canceling units to get the desired quantity. Hence,

$\text{748 }\cancel{\text{mm Hg}}\times \left(\frac{\frac{1}{760\text{ atm}}}{\text{ 1 }\cancel{\text{mm Hg}}}\right)=0.98\text{ atm}$

One degree Celsius $\left({}^{\text{°}}\text{C}\right)$ is equal to $\text{273}\text{K}$. It can be expressed as:

${\text{1}}^{\text{°}}\text{C}=\text{273}\text{K}$

Convert the temperature $\left(T\right)$ from ${}^{\text{°}}\text{C}$ and $\text{K}$ as:

$T={}^{\text{o}}\text{C}+273$

Substitute $27.0{}^{\text{°}}\text{C}$ in the above expression as:

$\begin{array}{c}{T}_1=27{}^{\text{°}}\text{C}+273\\ =300\text{K}\end{array}$

The ideal gas equation is expressed as follows:

$PV=nRT$

Rearrange the above expression for the calculation of n (number of moles) as follows:

$n=\frac{PV}{RT}$

Substitute $\text{0}\text{.98 atm}$ for $P$, $\text{300 K}$ for $T$, $0.0821\frac{\text{L }\text{. atm}}{\text{mol }\text{. K}}$ for $R$, and $\text{4}\text{.00 L}$ for $V$ in the above expression as follows:

$\begin{array}{c}n=\frac{\left(\text{0}\text{.98}\cancel{\text{ atm}}\right)\left(\text{4}\text{.00 }\cancel{\text{L}}\right)}{\left(0.0821\frac{\cancel{\text{L}}\text{. }\cancel{\text{atm}}}{\text{mol }\text{. }\cancel{\text{K}}}\right)\left(\text{300 }\cancel{\text{K}}\right)}\\ =0.160\text{ mol}\end{array}$

The mass of benzene (solvent) is $75\text{ g}$.

The conversion factor for $\text{g}$ to kilograms $\left(\text{kg}\right)$ is $\left(\frac{{10}^{-3}\text{ kg}}{\text{1 g}}\right)$.

Multiply the conversion factor with the given quantity and cancelling units to get the desired quantity. Hence,

$75\cancel{\text{g}}\times \frac{{10}^{-3}\text{ kg}}{\text{1 }\cancel{\text{g}}}=75\times {10}^{-3}\text{ kg}$

The molality is calculated as follows:

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

Substitute $\text{0}\text{.160 mol}$ for moles of solute and $75\times {10}^{-3}\text{ kg}$ for mass of solvent in the above expression as follows:

$\begin{array}{c}m=\frac{0.160\text{ mol }}{75\times {\text{10}}^{-3}\text{ kg }}\\ =2.13m\end{array}$

The freezing point depression is calculated as follows:

$\Delta T_{\text{f}}=m\times K_{\text{f}}$

Substitute $\text{2}\text{.13 }m$ for $m$ and $\text{5}\text{.12 }\frac{\text{C}}{m}$ for $K_{\text{f}}$ in the above expression as follows:

$\begin{array}{c}\Delta T_{\text{f}}=2.13\cancel{m}\times 5.12\frac{\text{C}}{\cancel{m}}\\ =10.9\text{C}\end{array}$

The freezing point of the solution is calculated as follows:

$T_f=T_f{}^{°}-\Delta T_{\text{f}}$

Substitute $\text{5}\text{.5 }\text{C}$ for $T_f{}^{°}$( freezing point of pure benzene is $\text{5}\text{.5 }\text{C}$) and $\text{10}\text{.9 }\text{C}$ for $\Delta T_{\text{f}}$ in the above expression as follows:

$\begin{array}{c}{T}_f\text{ = 5}\text{.5 }\text{C}-10.9\text{C}\\ \text{=}-5.4\text{C}\end{array}$

Conclusion

The freezing point of the solution is $-5.4{}^{°}\text{C}$.