Question

If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945
You may assume L = 1.20cm. Hint: A=εLC

b. Convert the molarity in part a to w/v%. Show your work. Molar mass of FD&C red #3 = 879.86g/mol Refer to examples on next page

c. If you want to dilute the red dye solution in part a by 5 times in a single dilution step, explain in two sentences on how one should proceed with the dilution. In addition, calculate the final concentration of the diluted solution. Show your work for the numerical part of this question.

Answer 1

Explanation:

a) Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

$\epsilon$ = molar absorptivity of this solution =$5.56\times 10^4 M^{-1} cm^{-1}$

$0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c$

$c=1.4163\times 10^{-5} M=14.16 \mu M$

($1\mu M=10^{-6} M$)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) $c=1.4163\times 10^{-5} mol/L$

1 L of solution contains $1.4163\times 10^{-5}$ moles of red dye.

Mass of $1.4163\times 10^{-5}$ moles of red dye:

$1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g$

$(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100$

$red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%$

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = $c=1.4163\times 10^{-5} M$

Concentration of red solution after dilution = c'

$c=c'\times 5$

$1.4163\times 10^{-5} M=c'\times 5$

$c'=2.83\times 10^{-6} M$

The final concentration of the diluted solution is $2.83\times 10^{-6} M$