Chapter 5, Problem 5A.1ST

Interpretation Introduction

Interpretation: Under the given conditions the partial molar volume of ethanol has to be calculated.

Concept introduction: The partial molar volume is the contribution that a component of mixture makes to the total volume of a sample.  The total volume of the mixture of two components $\text{A}$ and $\text{B}$ is given by expression,

  $V=n_{\text{A}}{V}_{\text{A}}+n_{\text{B}}{V}_{\text{B}}$

Answer to Problem 5A.1ST

The molar volume of ethanol is $\underset{\_}{56.67c{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The expression for the total volume in term of number of moles and partial molar volume of ethanol/water solution is shown as,

  $V=n_{\text{Ethanol}}{V}_{\text{Ethanol}}+n_{\text{Water}}{V}_{\text{Water}}$

Rearrange the above expression in term of $V_{\text{Ethanol}}$.

  $\begin{array}{c}V=n_{\text{Ethanol}}{V}_{\text{Ethanol}}+n_{\text{Water}}{V}_{\text{Water}}\\ n_{\text{Ethanol}}{V}_{\text{Ethanol}}=V-n_{\text{Water}}{V}_{\text{Water}}\\ V_{\text{Ethanol}}=\frac{V-n_{\text{Water}}{V}_{\text{Water}}}{n_{\text{Ethanol}}}\end{array}$                                                      (1)

Where,

• • $V_{\text{Ethanol}}$ is the molar volume of ethanol.
• • $V_{\text{Water}}$ is the molar volume of water.
• • $n_{\text{Ethanol}}$ is the number of moles of ethanol.
• • $n_{\text{Water}}$ is the number of moles of water.

It is given that,

Mass density is $0.914\text{g}{\text{cm}}^{-\text{3}}$.

Molar volume of water $\left(V_{\text{Water}}\right)$ is $17.4{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}$.

Since the given solution is $50$ percentage by mass ethanol/water.  That means $50\text{g}$ of ethanol is dissolved in $50\text{g}$ of water.  Thus, the total mass of the solution is, $100\text{g}$.

The total volume of solution is calculated as,

$V=\frac{M}{\rho }$                                                                                               (2)

Where,

• • $V$ is total volume of solution.
• • $M$ is the mass of solution.
• • $\rho$ is the mass density of solution.

Substitute the given values of mass and mass density in equation (2).

   $\begin{array}{c}V=\frac{M}{\rho }\\ =\frac{100\text{g}}{0.914\text{g}{\text{cm}}^{-3}}\\ =109.4{\text{cm}}^{\text{3}}\end{array}$

The number of moles of ethanol and water is calculated as,

  $n_{\text{Ethanol}}=\frac{m_{\text{Ethanol}}}{m{w}_{\text{Ethanol}}}$ (3)

  $n_{\text{Water}}=\frac{m_{\text{Water}}}{m{w}_{\text{Water}}}$                                                                                              (4)

Where,

• • $m_{\text{Ethanol}}$ is the mass of ethanol.
• • $m{w}_{\text{Ethanol}}$ is the molar mass of ethanol $\left(46.08\text{g}{\text{mol}}^{-\text{1}}\right)$.
• • $m_{\text{Water}}$ is the mass of water.
• • $m{w}_{\text{Water}}$ is the molar mass of water $\left(18.02\text{}\text{g}\text{}\text{}{\text{mol}}^{-1}\right)$

Substitute the given value of mass and molar mass in equation (3) and (4).

  $\begin{array}{c}{n}_{\text{Ethanol}}=\frac{\text{50}\text{g}}{\text{46}\text{.08}\text{g}{\text{mol}}^{-\text{1}}}\\ =1.08\text{mol}\end{array}$

  $\begin{array}{c}{n}_{\text{Water}}=\frac{\text{50}\text{g}}{\text{18}\text{.02}\text{g}{\text{mol}}^{-\text{1}}}\\ =2.77\text{mol}\end{array}$

Substitute the values of $V,n_{\text{Ethanol}},V_{\text{Water}}$ and $n_{\text{Water}}$ in equation (1).

  $\begin{array}{c}{V}_{\text{Ethanol}}=\frac{V-n_{\text{Water}}{V}_{\text{Water}}}{n_{\text{Ethanol}}}\\ =\left(\frac{\text{109}\text{.4}{\text{cm}}^{\text{3}}-\left(\text{2}\text{.77}\text{mol}\times \text{17}\text{.4}{\text{cm}}^{\text{3}}{\text{mol}}^{-\text{1}}\right)}{\text{1}\text{.08}\text{mol}}\right)\\ =\left(\frac{61.20}{1.08}\right){\text{cm}}^{\text{3}}{\text{mol}}^{-1}\\ =\underset{\_}{56.67c{m}^{3}mo{l}^{-1}}\end{array}$

Therefore, the molar volume of ethanol is $\underset{\_}{56.67c{m}^{3}mo{l}^{-1}}$.