Chapter 5, Problem 5A.2BE

Interpretation Introduction

Interpretation: The partial molar volume of the components $\text{A}$ and $\text{B}$, (where $\text{A}$ represents to water and $\text{B}$ represents to ${\text{MgSO}}_4$) in a solution has to be calculated.

Concept introduction: Mixtures of two components are called as binary mixture.  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,

  $V={\left(\frac{\partial V}{\partial n}\right)}_{p,T,n\text{'}}$

Answer to Problem 5A.2BE

The partial molar volumes of ${\text{MgSO}}_4$ and ${\text{H}}_{\text{2}}\text{O}$ are $\underset{\_}{-1.38c{m}^3\text{}mo{l}^{-1}}$ and $\underset{\_}{18.04c{m}^3\text{}mo{l}^{-1}}$ respectively.

Explanation of Solution

The partial molar volumes of components of mixture vary with compositions.  The nature of molecule changes from pure $\text{A}$ to pure $\text{B}$ with change in composition.  This results in the variation in thermodynamic properties of a mixture.  The partial molar volume of component $\text{B}$ is expressed as,

  $V_{\text{B}}={\left(\frac{\partial V}{\partial n_{\text{B}}}\right)}_{p,T,n_{\text{A}}}$                                                                                             (1)

Where,

• • $V_{\text{B}}$ is partial molar volume of component $\text{B}$.
• • $\partial V$ is the total volume of the mixture.
• • $n_{\text{B}}$ is the mole fraction of component $\text{B}$.

It is given that,

Weight of water $=1\text{kg}$.or $1000\text{g}$.

Since, molecular weight of water is $18\text{}\text{g}{\text{mol}}^{-\text{1}}$.

Therefore the number of moles of water is,

  $\begin{array}{c}{n}_{{\text{H}}_{\text{2}}\text{O}}=\frac{\text{Weight}}{\text{Molecular}\text{weight}}\\ =\frac{\text{1000}\text{g}}{\text{18}\text{g}{\text{mol}}^{-\text{1}}}\\ =55.5\text{mol}\end{array}$

Volume of an aqueous solution of ${\text{MgSO}}_4$ in water is,

  $v=\text{1001}\text{.21}+34.69{\left(x-0.070\right)}^2$                                                                      (2)

Where,

• • $v=\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{${\text{cm}}^{\text{3}}$}\right.$
• • $x=\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$b^{\oplus }$}\right.$

Molality of ${\text{MgSO}}_4=\text{0}\text{.050}\text{mol}{\text{kg}}^{-\text{1}}$

For pure form, $b^{\oplus }$ is taken as unity.

Hence,

  $x=\frac{n_{{\text{MgSO}}_4}}{\text{1}\text{mol}}$

The partial molar volume of ${\text{MgSO}}_4$ in the mixture is expressed as,

  $V_{{\text{MgSO}}_4}={\left(\frac{\partial V}{\partial n_{{\text{MgSO}}_4}}\right)}_{p,T,n_{\text{A}}}$                                                                                  (3)

Substitute the value of $V$ and $n_{{\text{MgSO}}_4}$ in equation (3).

  $\begin{array}{c}{V}_{{\text{MgSO}}_4}={\left(\frac{\partial V}{\partial n_{{\text{MgSO}}_4}}\right)}_{p,T,n_{\text{A}}}\\ ={\left(\frac{\partial \left(v{\text{cm}}^3\right)}{\partial \left(x\text{mol}\right)}\right)}_{p,T,n_{\text{A}}}\\ ={\left(\frac{\partial v}{\partial x}\right)}_{p,T,n_{\text{A}}}{\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\end{array}$

Substitute the value of $v$ from equation (2) in the above expression.

  $\begin{array}{c}{\left(\frac{\partial v}{\partial x}\right)}_{p,T,n_{\text{A}}}=\left(\frac{\partial \left(\text{1001}\text{.21}+34.69{\left(x-0.070\right)}^2\right)}{\partial x}\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =\left(\text{0}+\text{34}\text{.69}\times \text{2}\left(x-\text{0}\text{.070}\right)\times \text{1}\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =\left(69.38\left(x-0.070\right)\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =\left(69.38x-4.85\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\end{array}$

The partial molar volume of ${\text{MgSO}}_4$ in term of $x$ is,

  ${\left(\frac{\partial v}{\partial x}\right)}_{p,T,n_{\text{A}}}=\left(69.38x-4.85\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}$                                                          (4)

To calculate the partial molar volume of ${\text{MgSO}}_4$, substitute the value of $x$ in equation (4).

The number of moles of ${\text{MgSO}}_4$ is calculated by the formula

  ${\text{n}}_{{\text{MgSO}}_4}=\text{Molality}\times \text{Mass}$

The molality is $\text{0}\text{.050}\text{mol}{\text{kg}}^{-\text{1}}$.

The mass of the solution is $1\text{kg}$.

Substitute the value of molality and mass in the above equation

  $\begin{array}{c}{\text{n}}_{{\text{MgSO}}_4}=\text{Molality}\times \text{Volume}\\ =\text{0}\text{.050}\text{mol}{\text{kg}}^{-\text{1}}\times \text{1kg}\\ =\text{0}\text{.050}\text{mol}\end{array}$

Hence,

  $\begin{array}{c}x=\frac{0.050\text{mol}}{1\text{mol}}\\ =0.050\end{array}$

Substitute the value of $x$ in equation (4).

  $\begin{array}{c}{\left(\frac{\partial v}{\partial x}\right)}_{p,T,n_{\text{A}}}=\left(69.38x-4.85\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =\left(\text{69}\text{.38}\times \text{0}\text{.050}-4.85\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =\left(3.469-4.85\right){\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\\ =-1.38{\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}\end{array}$

Therefore, $V_{{\text{MgSO}}_4}=-1.38{\text{cm}}^{\text{3}}\text{}{\text{mol}}^{-\text{1}}$

Therefore, the partial molar volume of ${\text{MgSO}}_4$ is calculated as $\underset{\_}{-1.38c{m}^3\text{}mo{l}^{-1}}$.

Now the total volume of the mixture is given by the relation,

  $v=v_{\text{A}}{n}_{\text{A}}+v_{\text{B}}{n}_{\text{B}}$                                                                                               (5)

Where,

• • $v_{\text{A}}$ is the molar volume of water.
• • $v_{\text{B}}$ is the molar volume of ${\text{MgSO}}_4$.

Rearrange the above expression in term of $v_{\text{A}}$.

  $\begin{array}{c}v=v_{\text{A}}{n}_{\text{A}}+v_{\text{B}}{n}_{\text{B}}\\ v-v_{\text{B}}{n}_{\text{B}}=v_{\text{A}}{n}_{\text{A}}\end{array}$

Or,

    $v_{\text{A}}=\frac{v-v_{\text{B}}{n}_{\text{B}}}{n_{\text{A}}}$                                                                                                  (6)

The total volume $v$ is calculated by substituting the value of $x$ in equation (2).

  $\begin{array}{c}v=\text{1001}\text{.21}+34.69{\left(x-0.070\right)}^2{\text{cm}}^{\text{3}}\\ =\text{1001}\text{.21}+34.69{\left(0.050-0.070\right)}^2{\text{cm}}^{\text{3}}\\ =\text{1001}\text{.21}+34.69{\left(-0.020\right)}^2{\text{cm}}^{\text{3}}\\ =1001.22{\text{cm}}^{\text{3}}\end{array}$

Substitute the values of $v,v_{\text{B}},n_{\text{A}},n_{\text{B}}$ in equation (6).

  $\begin{array}{c}{v}_{\text{A}}=\frac{v-v_{\text{B}}{n}_{\text{B}}}{n_{\text{A}}}\\ =\frac{\left(\text{1001}\text{.22}-\left(-\text{1}\text{.38}\right)\times 0.\text{050}\right)}{55.5}{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}\\ =\left(\frac{1001.22+0.069}{55.5}\right){\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}\\ =18.04{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}\end{array}$

Therefore, the partial molar volume of ${\text{H}}_{\text{2}}\text{O}$ is calculated as, $\underset{\_}{18.04c{m}^3\text{}mo{l}^{-1}}$.