Chapter 5, Problem 5A.5P

Interpretation Introduction

Interpretation:

The partial molar volume of ${\text{CuSO}}_{\text{4}}$ in the range of measurements has to be determined and plotted.

Concept Introduction:

The partial molar volume is the contribution that a component of mixture makes to the total volume of a sample. The partial molar volume of a substance is expressed as,

    $V_{\text{j}}={\left(\frac{dV}{d{n}_{\text{j}}}\right)}_{p,T,n_{\text{w}}}$

Answer to Problem 5A.5P

The partial molar volume of ${\text{CuSO}}_{\text{4}}$ in the range of measurements is calculated in the table below.

$m_{\text{A}}/\text{g}$	$\rho /\text{g}{\text{cm}}^{-3}$	$x_{\text{A}}$	$V_{\text{j}}/{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}$
5	1.0151	0.006	-15.946
10	1.107	0.012	-10.357
15	1.167	0.0195	-3.37075
20	1.23	0.027	3.6155

The graph between mole fraction of ${\text{CuSO}}_{\text{4}}$, $x_{\text{A}}$ and partial molar volume $V_{\text{j}}$ is plotted below.

Figure 2

Explanation of Solution

The data of mass densities of aqueous solution of ${\text{CuSO}}_{\text{4}}$ is given in the table shown below.

$m\left({\text{CuSO}}_{\text{4}}\right)/\text{g}$	$\rho /\text{g}{\text{cm}}^{-3}$
5	1.0151
10	1.107
15	1.167
20	1.23

The moles of ${\text{CuSO}}_{\text{4}}$ is calculated by the formula given below.

    $\text{Moles}\text{of}{\text{CuSO}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{CuSO}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{CuSO}}_{\text{4}}}$

The molar mass of ${\text{CuSO}}_{\text{4}}$ is $159.61\text{g}{\text{mol}}^{-1}$.

Substitute the molar mass of ${\text{CuSO}}_{\text{4}}$ in the above equation.

    $\text{Moles}\text{of}{\text{CuSO}}_{\text{4}}=\frac{\text{Mass}\text{of}{\text{CuSO}}_{\text{4}}}{159.61\text{g}{\text{mol}}^{-1}}\left(1\right)$

The density of the solution ($\rho$) is calculated by the formula given below.

    $\begin{array}{c}\text{ρ=}\frac{\text{Mass}\text{of}\text{the}\text{solution}}{\text{Volume}\text{of}\text{the}\text{solution}}\\ \text{=}\frac{\text{m}{\text{A}}_{}{\text{+m}}_{\text{B}}}{\text{V}}\end{array}$

Where,

${\text{m}}_{\text{A}}$ is the mass of ${\text{CuSO}}_{\text{4}}$.

${\text{m}}_{\text{B}}$ is the mass of ${\text{H}}_{\text{2}}\text{O}$.

Therefore, the molar density of the solution (${\rho }_{\text{m}}$) is given by the formula below.

    ${\text{ρ}}_{\text{m}}\text{=}\frac{\frac{\text{m}{\text{A}}_{}}{{\text{M}}_{\text{A}}}\text{+}\frac{{\text{m}}_{\text{B}}}{{\text{M}}_{\text{B}}}}{\text{V}}\left(\text{2}\right)$

Where,

${\text{m}}_{\text{A}}$ is the mass of ${\text{CuSO}}_{\text{4}}$.

${\text{m}}_{\text{B}}$ is the mass of ${\text{H}}_{\text{2}}\text{O}$.

$V$ is the volume of the solution.

The volume of the solution is given by the formula below.

    $\text{V=}\frac{\text{Mass}\text{of}\text{solution}}{\text{ρ}}\left(\text{3}\right)$

The mass of the solution is $\text{100}\text{g}$.

Substitute the value of mass of solution in (3).

    $\text{V=}\frac{\text{100}}{\text{ρ}}$

Substitute the value of $V$ in equation (2).

    $\begin{array}{c}{\text{ρ}}_{\text{m}}\text{=}\frac{\frac{\text{m}{\text{A}}_{}}{{\text{M}}_{\text{A}}}\text{+}\frac{{\text{m}}_{\text{B}}}{{\text{M}}_{\text{B}}}}{\frac{\text{100}}{\text{ρ}}}\\ \text{=}\frac{\text{ρ}}{\text{100}}\left(\frac{\text{m}{\text{A}}_{}}{{\text{M}}_{\text{A}}}\text{+}\frac{{\text{m}}_{\text{B}}}{{\text{M}}_{\text{B}}}\right)\left(\text{4}\right)\end{array}$

The mass of water ($m_{\text{B}}$) can be written as $100-m_{\text{A}}$.

Substitute the value of $m_{\text{B}}$ in equation (4).

    ${\text{ρ}}_{\text{m}}\text{=}\frac{\text{ρ}}{\text{100}}\left(\frac{\text{m}{\text{A}}_{}}{{\text{M}}_{\text{A}}}\text{+}\frac{\left(\text{100-m}{\text{A}}_{}\right)}{{\text{M}}_{\text{B}}}\right)\left(\text{5}\right)$

The total molar volume of the solution ($V_{\text{m}}$) is expressed by the equation shown below.

    ${\text{V}}_{\text{m}}\text{=}\frac{\text{1}}{{\text{ρ}}_{\text{m}}}\left(\text{6}\right)$

Substitute the value of ${\rho }_{\text{m}}$ in equation (6).

    ${\text{V}}_{\text{m}}\text{=}\frac{\text{1}}{\frac{\text{ρ}}{\text{100}}\left(\frac{\text{m}{\text{A}}_{}}{{\text{M}}_{\text{A}}}\text{+}\frac{\text{100-m}{\text{A}}_{}}{{\text{M}}_{\text{B}}}\right)}\left(\text{7}\right)$

The mass of ${\text{CuSO}}_{\text{4}}$ ($m_{\text{A}}$) is $\text{5}\text{g}$.

The density ($\rho$) is $1.051\text{g}{\text{cm}}^{-3}$

The molar mass of ${\text{CuSO}}_{\text{4}}$ ($M_{\text{A}}$) is $159.61\text{g}{\text{mol}}^{-1}$.

The molar mass of ${\text{H}}_2\text{O}$ ($M_{\text{B}}$) is $18\text{g}{\text{mol}}^{-1}$.

Substitute the value of $m_{\text{A}}$, $\rho$, $M_{\text{A}}$, and $M_{\text{B}}$ in equation (7).

    $\begin{array}{c}{\text{V}}_{\text{m}}\text{=}\frac{\text{1}}{\frac{\text{1}\text{.051}\text{g}{\text{cm}}^{\text{-3}}}{\text{100}}\left(\frac{\text{5}\text{g}}{\text{159}\text{.61}\text{g}{\text{mol}}^{\text{-1}}}\text{+}\frac{\text{100-m}{\text{A}}_{}}{\text{18}\text{g}{\text{mol}}^{\text{-1}}}\right)}\\ \text{=17}\text{.9}{\text{cm}}^{\text{-3}}{\text{mol}}^{\text{-1}}\end{array}$

The mole fraction of ${\text{CuSO}}_{\text{4}}$ ($x_{\text{A}}$) is calculated by the formula below.

    $x_{\text{A}}=\frac{\frac{\text{Mass}\text{of}{\text{CuSO}}_{\text{4}}}{159.61\text{g}{\text{mol}}^{-1}}}{\frac{\text{Mass}\text{of}{\text{CuSO}}_{\text{4}}}{159.61\text{g}{\text{mol}}^{-1}}+\frac{\left(\text{100}-\text{Mass}\text{of}{\text{CuSO}}_{\text{4}}\right)}{18\text{g}{\text{mol}}^{-1}}}\left(8\right)$

The mass of ${\text{CuSO}}_{\text{4}}$ ($m_{\text{A}}$) is $\text{5}\text{g}$.

Substitute the mass of ${\text{CuSO}}_{\text{4}}$ in equation (8).

    $\begin{array}{c}{x}_{\text{A}}=\frac{\frac{\text{5}\text{g}}{159.61\text{g}{\text{mol}}^{-1}}}{\frac{\text{5}\text{g}}{159.61\text{g}{\text{mol}}^{-1}}+\frac{\left(\text{100}\text{g}-\text{5}\text{g}\right)}{18\text{g}{\text{mol}}^{-1}}}\\ =\frac{0.0313{\text{mol}}^{-1}}{0.0313{\text{mol}}^{-1}+5.2778{\text{mol}}^{-1}}\\ =0.006\end{array}$

Similarly the values of $x_{\text{A}}$ and $V_{\text{m}}$ are calculated for all measurements in the table below.

$m_{\text{A}}/\text{g}$	$\rho /\text{g}{\text{cm}}^{-3}$	$x_{\text{A}}$	$V_{\text{m}}$
5	1.0151	0.006	17.94
10	1.107	0.012	17.86
15	1.167	0.0195	17.81
20	1.23	0.027	17.81

The values of $x_{\text{A}}$ were plotted against $V_{\text{m}}$ as shown below.

Figure 1

The data was fit to a second order polynomial using the POLY program to generate the equation given below.

  $V_{\text{m}}/{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}=465.75x_{\text{A}}^2-21.535x_{\text{A}}+18.052\left(9\right)$

The partial molar volume of a substance ($V_j$) is expressed as,

    ${\text{V}}_{\text{j}}\text{=}{\left(\frac{{\text{dV}}_{\text{m}}}{{\text{dx}}_{\text{j}}}\right)}_{{\text{p,T,x}}_{\text{w}}}$

Where,

    $x_{\text{j}}$ is the mole fraction.

Differentiate equation (9) with respect to $x_{\text{A}}$.

    $\begin{array}{c}{\text{V}}_{\text{j}}\text{=}\frac{\text{d}{\text{V}}_{\text{m}}/{\text{cm}}^{\text{3}}{\text{mol}}^{\text{-1}}}{{\text{dx}}_{\text{A}}}\\ \text{=}\frac{\text{d}}{{\text{dx}}_{\text{A}}}\left(\text{465}{\text{.75x}}_{\text{A}}^{\text{2}}\text{-21}{\text{.535x}}_{\text{A}}\text{+18}\text{.052}\right)\\ \text{=931}{\text{.5x}}_{\text{A}}\text{-21}\text{.535}\left(\text{10}\right)\end{array}$

Substitute the value of $x_{\text{A}}=0.006$ in equation (10).

    $\begin{array}{c}\frac{\text{d}{V}_{\text{m}}/{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}}{\text{d}{x}_{\text{A}}}=931.5\times 0.006-21.535\\ =\underset{\_}{-15.946c{m}^{3}mo{l}^{-1}}\end{array}$

Similarly the values of $V_j$ is calculated all mole fractions $x_{\text{A}}$ in the table below.

$x_{\text{A}}$	$V_{\text{j}}/{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}$
0.006	-15.946
0.012	-10.357
0.0195	-3.37075
0.027	3.6155

The graph between mole fraction of ${\text{CuSO}}_{\text{4}}$, $x_{\text{A}}$ and partial molar volume $V_{\text{j}}$ is plotted below.

Figure 2