Chapter 14, Problem 14A.1ST

Interpretation Introduction

Interpretation:

The Z-average molar mass of given sample has to be estimated.

Concept Introduction:

The number-average molar mass, ${\text{M}}_{\text{n}}$, is obtained by weighting each molar mass by the number of molecules of that mass present in the sample:

    $\begin{array}{l}\overline{{\text{M}}_{\text{n}}}\text{=}\frac{\text{1}}{\text{N}}{\displaystyle \sum _{\text{i}}{\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}\\ \\ \overline{{\text{M}}_{\text{n}}}\text{is}\text{number-average molar mass}\\ \\ {\text{N}}_{\text{i}}{\text{is the number of molecules with molar mass M}}_{\text{i}}\end{array}$

Explanation of Solution

The Z-average molar mass, which is obtained in certain sedimentation experiments is calculated as,

    $\begin{array}{l}\overline{{\text{M}}_{\text{Z}}}\text{=}\frac{{\displaystyle \sum _{\text{i}}{\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}{}^{\text{3}}}{{\displaystyle \sum _{\text{i}}{\text{N}}_{\text{i}}}{\text{M}}_{\text{i}}{}^{\text{2}}}\\ \\ \text{=}\frac{\text{(1}\text{.3× 7}{\text{.5}}^{\text{3}}\text{ +}\text{0}\text{.70 ×12}{\text{.5}}^{\text{3}}\text{+}\text{ 0}\text{.51 ×17}{\text{.5}}^{\text{3}}\text{+}\text{0}\text{.25}\text{× 22}{\text{.5}}^{\text{3}}\text{ +}\text{0}\text{.11× 27}{\text{.5}}^{\text{3}}\text{+}\text{ 0}\text{.052 ×32}{\text{.5}}^{\text{3}}\text{)}}{\text{(1}\text{.3× 7}{\text{.5}}^{\text{2}}\text{ +}\text{0}\text{.70 ×12}{\text{.5}}^{\text{2}}\text{+}\text{ 0}\text{.51 ×17}{\text{.5}}^{\text{2}}\text{+}\text{0}\text{.25}\text{× 22}{\text{.5}}^{\text{2}}\text{ +}\text{0}\text{.11× 27}{\text{.5}}^{\text{2}}\text{+}\text{ 0}\text{.052 ×32}{\text{.5}}^{\text{2}}\text{)}}\\ \\ \text{=}{\text{19 kg mol}}^{\text{-1}}\end{array}$

Hence, Z-average molar mass of given sample is ${\text{19 kg mol}}^{\text{-1}}$