A solution is prepared by dissolving $0\mathrm{.}1093$ grams of FD&C Red #$40$ in water to produce $2\mathrm{.}000$ Liters of solution. Given that the molecular formula of Red $40$ is

$C_{18}{H}_{14}{N}_2{S}_2{O}_8,$ determine the molarity of this stock solution. $($Note: some sources will give a different formula for Red $40$ ; be sure to use this formula for your calculations$).$

$\frac{0\mathrm{.}1093}{450}=\frac{0\mathrm{.}00024289}{2\mathrm{.}000C}=0\mathrm{.}00012144M$

$25\mathrm{.}00mL$ of the dark red stock solution described question #$1$ is transferred to a volumetric flask and diluted with water to a total final volume of $200\mathrm{.}00mL.$

a$.$ how many moles of Red $40$ are in the $25\mathrm{.}00mL$ sample that was transferred?

$a.$ how many moles $of$ Red $40$ are $in$ the $25\mathrm{.}00mL$ sample that was transferred?

$(0\mathrm{.}025)(0\mathrm{.}00\mathrm{.}012184)=0\mathrm{.}00003036$

moles

b$.$ After the dilution, what is the molarity of the Red $49$ in the $200\mathrm{.}00mL$ of solution?

$\frac{0\mathrm{.}000003036}{0\mathrm{.}3}=0\mathrm{.}00004518M$

$6.2$

note: the preceding problem $($ #$2$a and $b)$ is a dilution calculation. See the last part of $M$ the discussion section, just before the procedure, for a discussion of this calculation. We will be doing dilution calculations in almost every experiment in this course.

$3.$ The solution described in problem #$2$ is placed in the spectrophotometer. Based on the calibration curve shown on the next page, what would you predict the absorbance to be$?$ Note: like the example given in the discussion, the graph. Notice that the slope and intercept of the calibration curve are provided in the figure on the next page.

Absorbance $=$

The following calibration curve was produced for Red $40.$ The slope was calculated to be $2\mathrm{.}1033{10}^4.$ The