Chapter 5, Problem 5.27P

a)

Interpretation Introduction

Interpretation:

The allilin equivalent was in the unspiked garlic has to be graphed and $\text{95\%}$ confidence interval has to be found out.

Concept Introduction:

Standard additions to one solution:

The graph can be drawn using the equation

$\text{plot}\underset{\text{Function}\text{to}\text{plot}\text{on}\text{y-axis}}{\underbrace{{\text{I}}_{\text{S+X}}\left(\frac{\text{V}}{{\text{V}}_{\text{0}}}\right)}}\text{versus }{\text{I}}_{\text{X}}\text{+}\frac{{\text{I}}_{\text{X}}}{{\left[\text{X}\right]}_{\text{i}}}\underset{\text{function}\text{to}\text{plot}\text{on}\text{x-axis}}{\underbrace{{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{{\text{V}}_{\text{0}}}\right)}}$

Where X-intercept is ${\left[\text{X}\right]}_{\text{i}}$

b)

Interpretation Introduction

Interpretation:

The allicin content of garlic $\left(\text{mg}\text{allicin/g}\text{garlic}\right)$ when $\text{2}\text{mol}$ of alliin are converted to $\text{1}\text{mol}$ have to be calculated.