Chapter 22, Problem 22.51SP

(a)

Interpretation Introduction

Interpretation:

For the given solution the values of $\overline{{\text{M}}_{\text{w}}}$ and $\overline{{\text{M}}_{\text{n}}}$ has to be calculated.

Concept Introduction:

Number average molar mass:

Number average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Weight average molar mass:

Weight average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Answer to Problem 22.51SP

The number average molar mass of the given solution is $\text{3}\text{.6}\text{kg/mol}$.

The weight average molar mass of the given solution is $\text{4}\text{.3}\text{kg/mol}$.

Explanation of Solution

Given information,

  Five molecules with molar masses $\text{1}\text{.0, 3}\text{.0, 4}\text{.0, 4}\text{.0 and 6}\text{.0}\text{kg/mol}$.

Calculate the number average molar mass

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{=}\frac{\left[\left(\text{1}\right)\left(\text{1}\text{.0}\right)\text{+}\left(\text{1}\right)\left(\text{3}\text{.0}\right)\text{+}\left(\text{2}\right)\left(\text{4}\text{.0}\right)\text{+}\left(\text{1}\right)\left(\text{6}\text{.0}\right)\right]\text{kg/mol}}{\text{5}}\text{=}\frac{\text{18}}{\text{5}}\text{kg/mol}\\ \\ \text{=}\text{3}\text{.6}\text{kg/mol}\end{array}$

The number average molar mass of the given solution is $\text{3}\text{.6}\text{kg/mol}$.

Calculate the weight average molar mass

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{=}\frac{\left[\left(\text{1}\right){\left(\text{1}\text{.0}\right)}^{\text{2}}\text{+}\left(\text{1}\right){\left(\text{3}\text{.0}\right)}^{\text{2}}\text{+}\left(\text{2}\right){\left(\text{4}\text{.0}\right)}^{\text{2}}\text{+}\left(\text{1}\right){\left(\text{6}\text{.0}\right)}^{\text{2}}\right]\text{kg/mol}}{\left(\text{1}\right)\left(\text{1}\text{.0}\right)\text{+}\left(\text{1}\right)\left(\text{3}\text{.0}\right)\text{+}\left(\text{2}\right)\left(\text{4}\text{.0}\right)\text{+}\left(\text{1}\right)\left(\text{6}\text{.0}\right)}\text{=}\frac{\text{78}}{\text{18}}\text{kg/mol}\\ \\ \text{=}\text{4}\text{.3}\text{kg/mol}\end{array}$

The weight average molar mass of the given solution is $\text{4}\text{.3}\text{kg/mol}$.

(b)

Interpretation Introduction

Interpretation:

The number average molar mass is equal to weight average molar mass when all the molecules have identical molar mass, this has to be explained.

Concept Introduction:

Number average molar mass:

Number average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Weight average molar mass:

Weight average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Explanation of Solution

Given information,

  All the five molecules have identical molar masses $\text{1}\text{.0 kg/mol}$.

Calculate the number average molar mass

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{=}\frac{\left(\text{4}\right)\left(\text{5}\text{kg/mol}\right)}{\text{4}}\text{=}\frac{\text{20}}{\text{4}}\text{kg/mol}\\ \\ \text{=}\text{5}\text{kg/mol}\end{array}$

The number average molar mass of the given solution is $\text{5}\text{kg/mol}$.

Calculate the weight average molar mass

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{=}\frac{\left(\text{4}\right){\left(\text{5}\text{kg/mol}\right)}^{\text{2}}}{\left(\text{4}\right)\left(\text{5}\text{kg/mol}\right)}\text{=}\frac{\text{100}}{\text{20}}\text{kg/mol}\\ \\ \text{=}\text{5}\text{kg/mol}\end{array}$

The weight average molar mass of the given solution is $\text{5}\text{kg/mol}$.

(c)

Interpretation Introduction

Interpretation:

Explanation for the comparison of weight and number average molar masses gives us information about the distribution of the size of synthetic polymers like polyethylene and poly(vinyl chloride) has to be given.

Concept Introduction:

Number average molar mass:

Number average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Weight average molar mass:

Weight average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Explanation of Solution

Given information,

Polyethylene and poly(vinyl chloride) is given as example.

The number and weight average molar mass values will be closer when a small spread in the distribution of sizes of polymer chains in the sample.

On the other hand, the number and weight average molar mass values will have larger difference when a large spread in the distribution of sizes of polymer chains in the sample.

(d)

Interpretation Introduction

Interpretation:

Myoglobin and cytochrome c have the same number and weight average molar mass values, but this is not the case for hemoglobin this has to be explained.

Concept Introduction:

Number average molar mass:

Number average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Weight average molar mass:

Weight average molar mass $\left(\overline{{\text{M}}_{\text{n}}}\right)$ is given by

  $\begin{array}{l}\overline{{\text{M}}_{\text{w}}}\text{=}\frac{{\displaystyle \sum {\text{N}}_{\text{i}}{\text{M}}_{\text{i}}{}^{\text{2}}}}{{\displaystyle \sum {\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}}\\ \\ \text{where,}\\ {\text{N}}_{\text{i}}\text{-}\text{number}\text{of}\text{molecules}\text{with}\text{molar}\text{mass}{\text{M}}_{\text{i}}\text{.}\end{array}$

Explanation of Solution

Given information,

  Myoglobin and cytochrome c have the same number and weight average molar mass values, but this is not true in the case for haemoglobin.

Myoglobin and cytochrome c have no subunits so there are no dissociations and hence both have equal number and weight average molar mass values $\left(\text{i}\text{.e}{\overline{\text{M}}}_{\text{n}}\text{=}{\overline{\text{M}}}_{\text{w}}\right)$.

On the other hand, haemoglobin molecule has four subunits and in solution those molecules will dissociate to a varying extent so there is a distribution of molar mass $\left(\text{i}\text{.e}{\overline{\text{M}}}_{\text{n}}\ne {\overline{\text{M}}}_{\text{w}}\right)$.