Chapter 23, Problem 6STP

Interpretation Introduction

Interpretation:

The mass of 1 mol of Sample I needs to be determined.

Concept introduction:

DNA stands for deoxyribonucleic acid and RNA stands for ribonucleic acid. DNA includes four bases such as adenine (A), guanine (G), cytosine (C), and thymine (T). RNA consists of four bases: adenine, cytosine, uracil, and guanine.

Answer to Problem 6STP

Option D is the correct option. $\text{2}{\text{.72×10}}^{\text{5}}\text{g}$ is the mass of 1 mol of Sample I.

Explanation of Solution

To determine the mass of the sample, we use the following formula.

${\text{M}}_{\text{1}}\text{=}{\text{M}}_{\text{A}}{\text{N}}_{\text{A}}\text{+}{\text{M}}_{\text{G}}{\text{N}}_{\text{G}}\text{+}{\text{M}}_{\text{C}}{\text{N}}_{\text{C}}\text{+}{\text{M}}_{\text{T}}{\text{N}}_{\text{T}}$

Here, ${\text{M}}_{\text{1}}$ is the mass of the sample-I.

${\text{M}}_{\text{A}}{\text{N}}_{\text{A}}$ is the mass of adenine multiplied by number of adenine.

${\text{M}}_{\text{G}}{\text{N}}_{\text{G}}$ is the mass of guanine multiplied by number of guanine.

${\text{M}}_{\text{C}}{\text{N}}_{\text{C}}$ is the mass of cytosine multiplied by number of cytosine.

${\text{M}}_{\text{T}}{\text{N}}_{\text{T}}$ is the mass of thymine multiplied by number of thymine.

Given, A, C, G and T have molar masses 347.22 g/mol, 323.20 g/mol, 363.23 g/mol and 338.21 g/mol respectively.

The given table is,

From the table, we have the number of Guanine $\left({\text{N}}_{\text{G}}\right)$ = 231 mol and number of thymine $\left({\text{N}}_{\text{T}}\right)$ = number of adenine $\left({\text{N}}_{\text{A}}\right)$ = 165 mol.

$\begin{array}{c}{\text{M}}_{\text{1}}\text{=}{\text{M}}_{\text{A}}{\text{N}}_{\text{A}}\text{+}{\text{M}}_{\text{G}}{\text{N}}_{\text{G}}\text{+}{\text{M}}_{\text{C}}{\text{N}}_{\text{C}}\text{+}{\text{M}}_{\text{T}}{\text{N}}_{\text{T}}\\ \text{=}\left(\text{347}\text{.22}\text{g/mol×165}\text{mol}\right)\text{+}\left(\text{363}\text{.23}\text{g/mol}\text{×231}\text{mol}\right)\text{+}\left(\text{323}\text{.20}\text{g/mol×231}\text{mol}\right)\text{+}\left(\text{338}\text{.21}\text{g/mol×165}\text{mol}\right)\\ \text{=}\text{2}{\text{.72×10}}^{\text{5}}\text{g}\end{array}$

Conclusion

The mass of 1 mol of Sample I is $\text{2}{\text{.72×10}}^{\text{5}}\text{g}$