Name I. A solution is prepared by dissolving 0.1896 produce 2.500 Liters of solu

Question

Name I. A solution is prepared by dissolving 0.1896 produce 2.500 Liters of solution. Given thatFDe grams of FD&C; Red #40 in water to the molecular formula of Red 40 is the molarity of this stock solution. Cul114N2S2On determine the molarityofthisse molecularforcRed#40 2. 25.00 mL of the dark red stock solution described volumetric flask and diluted with water to 100.00 mL. question #1 is transferredtoa moles of Red 40 are in the 25.00 ml, sample that was transferred? moles b. After the dilution, what is the molarity of the Red 40 in the 100.00 ml of solution? note: the preceding problem (#2) is a dilution calculation. A comement ormula for calculating concentrations afier dilution is M,V, - M2V2, where Mi and Vi are the molarity and volume of the concentrated solution, respectively, and M and V are molarity and volume of the dilute solution the 3. The solution described in problem #2 is placed in the spectrophotometer calnbration curve shown on the next page, what would you predict the abso Based on dto e the exam not just estimated from the curve are provided in the figure on the next page. ple given in the discussion, the absorbance should be calculated, graph. Notice that the slope and intercept of the calibration Absorbance-

Explanation / Answer

1.

Molecular formula of Red 40 = C18H14N2S2O8

Molar mass of C18H14N2S2O8 = 450 g/mol

So, 450 g of C18H14N2S2O8 = 1 mol
or, 1 g of C18H14N2S2O8 = (1/450) mol
or, 0.1896 g of C18H14N2S2O8 = (0.1896/450) mol = 0.0004 mol

Volume = 2.500 L

Molarity = Moles / Liter
             = (0.0004 mol) / (2.5 L)
             = 0.00016 M

2.

(a)

Molarity = 0.00016 M,
Volume = 25 mL = 0.025 L

Molarity = Moles / Liter
Moles = (0.00016 M) x (0.025 L)
            = 4.0 x 10-6 moles

(b)

We know that

M1V1 = M2V2

So,

(0.00016 M) x (25 mL) = M2 x 100 mL
M2 = (0.00016 M) x (25 mL) / 100 mL
      = 4.0 x 10-5 M

3.

Slope = 2.1033 x 104

Intercept = 0.0039

We know that

Abs = e c l

where
e is the molar extinction coefficient (mol-1 dm3 cm-1),
c is the concentration (mol dm-3) and
l is the pathlength (cm).

Again, the plot of abs vs concentration gives a slope = 2.1033 x 104

Abs = (e l) c

So, e l = 2.1033 x 104

C = concentration from problem 2 = 4.0 x 10-5 M

So, Abs = (2.1033 x 104) x (4.0 x 10-5) = 0.84

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