ABSORBANCE ETRY: ADVANCE STUDY ASSIGNMENT Name 1. A solution grams FD&c; Red #40
ID: 532358 • Letter: A
Question
ABSORBANCE ETRY: ADVANCE STUDY ASSIGNMENT Name 1. A solution grams FD&c; Red #40 in water to produce is prepared by di of 40 is 2.500 Given that the molecular formula ofRed Ci Hi N2Szos, determine the mol of this stock solution. 2. 25.00 mL 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 100.00 mL. a. how many moles of Red 40 are in the 25.00 ml ple that was transferred? moles b. After the dilution, what is the molarity of the Red 40 in the 100.00 mLof solution? note: the preceding problem (H2) is a dilution calculation. A convenient formula for calculating concentrations after dilution is Mrvi-M,va where and Vi are the molarity and volume of the concentrated solution, respectively, and Mr and y, are the molarity and volume of the dilute solution. 3. a. The solution described in problem #2 is placed in the spectropher Ba on be? Note: like the example given in the discussion, the absorb should be calculated. ance not just from the graph. that the slope and intercept of the calibration curve are provided in the figure on the next page. AbsorbanceExplanation / Answer
Solved first two problems with multiple sub-parts as per chegg guidelines,post multiple question to get the remaining answers
1)
Molar mass of Red40 (C18H14N2S2O8) = 12 * 18 + 1 * 14 + 2 * 14 + 2 * 32 + 8 * 16 = 450 gm/mol
number of moles of red40 in solution = mass/molar mass = 0.1896/450 = 4.213 * 10^(-4)
molarity of the stock solution = Number of moles of solute/volume of solution in L
=> 4.213 * 10^(-4)/2.50 = 1.6853 * 10^(-4) M
2)
a) number of moles in 25 mL sample = 4.213 * 10^(-4) * 25mL/2500 mL = 4.213 * 10^(-6) moles
b) M1V1 = M2V2
1.6853 * 10^(-4) * 25 = M2 * 100
M2 = 4.21333 * 10^(-5) M
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