91Ó°ÊÓ

$$\begin{gathered} •\mathrm{C}3\mathrm{H}5\mathrm{IO}\mathrm{is}\mathrm{the}\mathrm{chemical}\mathrm{formula}\mathrm{for}\mathrm{iodoacetone},\mathrm{an}\mathrm{organoiodine}\mathrm{compound}. \\ •\mathrm{Under}\mathrm{normal}\mathrm{circumstances},\mathrm{the}\mathrm{material}\mathrm{is}\mathrm{a}\mathrm{white}\mathrm{liquid}\mathrm{that}\mathrm{is}\mathrm{soluble}\mathrm{in}\mathrm{ethanol}. \end{gathered}$$

It is necessary to calculate the concentration of lodoacetone in 3.00ml of the original solution.

The formula for quantitative analysis with internal standard can be found here,

$\frac{{\mathrm{A}}_{\mathrm{x}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{{\mathrm{A}}_{\mathrm{s}}}{\left[\mathrm{S}\right]}$

Here ${\mathrm{A}}_{\mathrm{x}}=$signal region of analyte

${\mathrm{A}}_{\mathrm{s}}$=area of internal standard

[X] = analyte's concentration

[S]= concentration of internal standard

F= response factor

To figure out how much lodoacetone is in 3.00ml of the original solution

Given,

Lodoacetone molarity $=6.3×{10}^{-8}\mathrm{M}$

P-dichlorobenzene molarity $=2.0×{10}^{-7}\mathrm{M}\mathrm{Peak}\mathrm{areas}=395\&787$

Unknown volume of lodoacetone solutionrole="math" localid="1654853858008" $=3.00ml$

p-dichlorobenzene Volume and Molarity$=0.100\mathrm{ml}\&1.6×{10}^{-5}\mathrm{M}$

Lodoacetone has a peak area of 633 square metres.

p-dichlorobenzene has a peak area of =520.

The response factor is determined as follows,

$$\begin{gathered} \frac{395}{\left[63nM\right]}=\mathrm{F}\left(\frac{787}{\left[200\mathrm{nM}\right]}\right) \\ \mathrm{F}=1.59 \end{gathered}$$

When an internal standard is coupled with an unknown, the concentration is determined as,

$$\begin{gathered} \frac{0.100\mathrm{ml}}{10.00\mathrm{ml}}\left(1.6×{10}^{-5}\mathrm{M}\right)=0.16\mathrm{μ²Ñ} \\ \mathrm{Lodoacetone}\mathrm{concentration}\mathrm{is}\mathrm{determined}\mathrm{as}, \\ \frac{633}{\left[\mathrm{iodoacentone}\right]}=1.59\left(\frac{520}{0.16\mathrm{μ²Ñ}}\right) \\ \left[\mathrm{iodoacentone}\right]=0.122\mathrm{μ²Ñ} \\ \mathrm{In}\mathrm{the}\mathrm{original}\mathrm{unknown}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoacetone}\mathrm{is}\mathrm{determined}\mathrm{as}, \\ \left[\mathrm{iodoacentone}\right]=\frac{10.00}{3.00}\left(-0.122\mathrm{μ²Ñ}\right) \\ \left[\mathrm{iodoacentone}\right]=0.41\mathrm{μ²Ñ} \\ \mathrm{In}3.00\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoactone}=0.41\mathrm{μ²Ñ} \\ \mathrm{The}3.00\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{original}\mathrm{solution},\mathrm{the}\mathrm{concentration}\mathrm{of}\mathrm{lodoacetone}\mathrm{was} \\ \mathrm{determined}\mathrm{and}\mathrm{found}\mathrm{to}\mathrm{be}0.41\mathrm{μ²Ñ} \end{gathered}$$