3a. If the molar absorptivity constant for the red dye solution is 5.56×104 M-1

Question

3a. If the molar absorptivity constant for the red dye solution is 5.56×104 M-1 cm-1 , calculate the molarity of the red dye solution with absorption = 0.900. You may assume L = 1.22cm. Hint: A=LC

b. Convert the molarity in part a to w/v%. Show your work. Molar mass of FD&C red #3 = 879.86g/mol w/v% is defined as weight (in grams) of solute/100 mL of solution You should remember that w/v% always refers to the grams of materials in 100mL of solution. ppm is defined as mg of solute/ L of solution (1mg = 10-3 g) ppb is defined as µg of solute / L of solution (1µg = 10-6 g)

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, what is the final concentration of the diluted solution? Show your work for the numerical part of this question. You are only allowed to use a 5-mL volumetric pipet, a 10-mL volumetric pipet, a 50-mL volumetric flask and/or a 100-mL volumetric flask during the dilution. You may NOT reuse the same pipet during the dilution.

Explanation / Answer

3a. If the molar absorptivity constant for the red dye solution is 5.56×104 M-1 cm-1 , calculate the molarity of the red dye solution with absorption = 0.900. You may assume L = 1.22cm. Hint: A=LC

Solution :-

Given data

Pathlength L = 1.22 cm , = 5.56*10^4 M-1 cm-1

Absorbance A = 0.900

Molarity = ?

A=LC

Lets put the values in the formula

0.900 = 5.56*10^4 M-1 cm-1 * 1.22 cm * C

0.900 / 5.56*10^4 M-1 cm-1 * 1.22 cm = C

7.72*10^-4 M= C

Therefore the molarity of the Red dye = 7.72*10^-4 M

b. Convert the molarity in part a to w/v%. Show your work. Molar mass of FD&C red #3 = 879.86g/mol w/v% is defined as weight (in grams) of solute/100 mL of solution You should remember that w/v% always refers to the grams of materials in 100mL of solution. ppm is defined as mg of solute/ L of solution (1mg = 10-3 g) ppb is defined as µg of solute / L of solution (1µg = 10-6 g)

solution :-

lets assume we have 100 ml of solution

now lets calculate the moles of the red dye in the 100 ml solution

moles = molarity * volume in liter

moles of red dye = 7.72*10^-4 mol per L * 0.100 L = 7.72*10^-5 mol red dye

now lets calculate the mass of the red dye

mass = moles * molar mass

mass of red dye = 7.72*10^-5 mol * 879.86 g per mol = 0.068 g

now lets calculate the percent of the red dye

% (w/V) red dye = (0.068 g /100 ml) *100 % = 0.068 %

Therefore w/v percent of the red dye = 0.068 %

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, what is the final concentration of the diluted solution? Show your work for the numerical part of this question. You are only allowed to use a 5-mL volumetric pipet, a 10-mL volumetric pipet, a 50-mL volumetric flask and/or a 100-mL volumetric flask during the dilution. You may NOT reuse the same pipet during the dilution.

Solution :-

To dilute the red dye solution by 5 times we need to use the water to make the volume of the final solution 5 times the initial volume of the solution.

Initial concentration of the red dye M1 =7.72*10^-4 M

Initial volume V1 = 10 ml

Final volume V2 = 10 * 5 = 50 ml

Final concentration M2 = ?

M1V1=M2V2

Lets put the values in the formula

7.72*10^-4* 10.0 ml = M 2 * 50.0 ml

7.72*10^-4* 10.0 ml/ 50.0 ml = M2

1.54*10^-4 M = M2

Therefore the molarity of the diluted solution = 1.54*10^-4 M

So the procedure steps to achieve the dilution is as follows

Pipette out 10 ml of the orginal red dye solution using the 10 ,ml pipette and transfer it to the 50 ml volumetric flask and dilute up to the mark by adding water.

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