Chapter 5, Problem 5B.9P

Interpretation Introduction

Interpretation:

The molar mass of polychloroprene and its second osmotic virial coefficient has to be calculated.

Concept Introduction:

When large molecules dissolve to produce solutions that are not ideal solutions, it is assumed that the Van’t Hoff equation is the first term of the virial like expansion.  The osmotic virial equation is,

    $\Pi =\left[\text{J}\right]RT\left(1+B\left[\text{J}\right]+\mathrm{......}\right)$

Answer to Problem 5B.9P

The molar mass of polychloroprene  is $\underset{\_}{1.25\times 10^{5}gmo{l}^{-1}}$.  The value of second osmotic virial coefficient is $\underset{\_}{1.23\times 10^{4}d{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The equation used to determine the molar mass from osmotic pressure is,

    $\frac{\text{Π}}{{\text{c}}_{\text{mass,}\text{J}}}\text{=}\frac{\text{RT}}{\text{M}}\text{+}\left(\frac{\text{BRT}}{{\text{M}}^{\text{2}}}\right){\text{c}}_{\text{mass,}\text{J}}\left(\text{1}\right)$

Where,

$c_{\text{mass,}\text{J}}$ is the mass concentration of polychloroprene.

$\Pi$ is the osmotic pressure.

$R$ is the gas constant.

$T$ is the temperature.

$M$ is the molar mass.

The given data is,

$\Pi \left(\text{N}{\text{m}}^{-2}\right)$	$c_{\text{mass,}\text{J}}\left(\text{mg}{\text{cm}}^{-3}\right)$	$\frac{\Pi }{c_{\text{mass,}\text{J}}}\left(\text{N}{\text{m}}^{-2}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)$
$30$	$1.33$	$22.5$
$51$	$2.10$	$24.28$
$132$	$4.52$	$29.20$
$246$	$7.18$	$34.26$
$390$	$9.87$	$39.51$

The plot of $\frac{\Pi }{c_{\text{mass,}\text{J}}}$ versus $c_{\text{mass,}\text{J}}$ is shown below.

Figure 1

From the plot intercept is $20.09\text{N}{\text{m}}^{-2}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}$ and the slope is $1.974\text{N}{\text{m}}^{-2}{\left({\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)}^2$.

The intercept of the plot of $\frac{\Pi }{c_{\text{mass,}\text{J}}}$ versus $c_{\text{mass,}\text{J}}$ is given by the expression $\frac{RT}{M}$. Hence,

    $\frac{RT}{M}=20.09\text{N}{\text{m}}^{-2}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}\left(2\right)$

The value of $T$ is $30°\text{C}$.

The conversion of temperature, $T$ from degrees to Kelvin is shown below.

    $\begin{array}{c}\text{K}=°\text{C}+\text{273}\\ =30°\text{C}+273\\ =303\text{K}\end{array}$

The value of $T$ is $303\text{K}$.

The value of $R$ is $8.314\times {10}^6{\text{cm}}^{\text{3}}\text{N}{\text{m}}^{-2}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The conversion of $\text{mg}$ to $\text{g}$ is done as,

    $\begin{array}{c}1\text{mg}\text{=}{10}^{-3}\text{g}\\ 20.09\text{N}{\text{m}}^{-2}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}=20.09\text{N}\frac{{\text{m}}^{-2}}{\text{mg}}\times \frac{\text{1}\text{mg}}{{10}^{-3}\text{g}}{\text{cm}}^{\text{3}}\\ =20.09\text{N}\frac{{\text{m}}^{-2}}{\text{g}}\times {10}^3{\text{cm}}^{\text{3}}\end{array}$

Substitute the values of $T$, $R$ and intercept in equation (2).

    $\begin{array}{c}\frac{\left(8.314\times {10}^6{\text{cm}}^{\text{3}}\text{N}{\text{m}}^{-2}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(303\text{K}\right)}{M}=20.09\times {10}^3\text{N}{\text{m}}^{-2}{\text{g}}^{-1}{\text{cm}}^{\text{3}}\\ M=\frac{\left(8.314\times {10}^6{\text{cm}}^{\text{3}}\text{N}{\text{m}}^{-2}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(303\text{K}\right)}{20.09\times {10}^3\text{N}{\text{m}}^{-2}{\text{g}}^{-1}{\text{cm}}^{\text{3}}}\\ =\underset{\_}{1.25\times 10^{5}gmo{l}^{-1}}\end{array}$

The molar mass of polychloroprene  is $\underset{\_}{1.25\times 10^{5}gmo{l}^{-1}}$.

The slope of the plot of $\frac{\Pi }{c_{\text{mass,}\text{J}}}$ versus $c_{\text{mass,}\text{J}}$ is given by the expression $\frac{BRT}{M^2}$. The value of the slope is $1.974\text{N}{\text{m}}^{-2}{\left({\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)}^2$. Hence,

    $\begin{array}{c}\frac{BRT}{M^2}=1.974\text{N}{\text{m}}^{-2}{\left({\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)}^2\\ B=\frac{M}{\left(\frac{RT}{M}\right)}1.974\text{N}{\text{m}}^{-2}{\left({\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)}^2\end{array}$

Substitute the value of $\text{M}$ and $\frac{\text{RT}}{\text{M}}$ in the above expression.

    $\begin{array}{c}B=\frac{1.25\times {10}^5\text{g}{\text{mol}}^{-1}}{20.09\text{N}{\text{m}}^{-2}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}}1.974\text{N}{\text{m}}^{-2}{\left({\text{mg}}^{-1}{\text{cm}}^{\text{3}}\right)}^2\\ =1.23\times {10}^4\text{g}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}\end{array}$

The conversion of $\text{mg}$ to $\text{g}$ and ${\text{cm}}^{\text{3}}$ to ${\text{dm}}^{\text{3}}$ is done as,

    $\begin{array}{l}1\text{mg}\text{=}{10}^{-3}\text{g}\\ \text{1}{\text{cm}}^3={10}^{-3}{\text{dm}}^{\text{3}}\end{array}$

Hence, the conversion of $1.23\times {10}^4\text{g}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}$ to $1.23\times {10}^4{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$ is,

    $\begin{array}{c}1.23\times {10}^4\text{g}{\text{mg}}^{-1}{\text{cm}}^{\text{3}}{\text{mol}}^{-1}=1.23\times {10}^4\text{g}{\text{mol}}^{-1}\times \frac{{10}^{-3}{\text{dm}}^{\text{3}}}{1{\text{cm}}^{\text{3}}}\times \frac{1\text{mg}}{{10}^{-3}\text{g}}\\ =\underset{\_}{1.23\times 10^{4}d{m}^{3}mo{l}^{-1}}\end{array}$

The value of second osmotic virial coefficient is $\underset{\_}{1.23\times 10^{4}d{m}^{3}mo{l}^{-1}}$.