Please help solve the three problems below. The correct answer is to the right o

Question

Please help solve the three problems below. The correct answer is to the right of the problem. Please clearly show all work needed to solve.

tion will consume Ag ions and cause the equilibrium to shift toward the solid AgCi sample t 6. Suppose that lo 20.0 mW/cm2 of the light that passes through the sample ()? or a sample that has an absorbance of 0.5. What is the 7. To create the calibration curve mh cm a ion curve a 5.00 cm pathlength cuvette was used in a spectrophotometer Th of of a series of standard solutions. The wavelength used was 350 nm. a linear function with the equation of y 204.5x+0.1. If a sample aosorbance the same species with an unknown concentration is poure d into a 1.00 cm pathlength cuvette resulting absorbance was 0.175, what would be the molarity of the analyte in't 8. 0.12 grams of acetylsalicylic acid (MW-180.157 g/mol) was heated with 1M NaOH an unknown sample? 1.33 X10-3m was heated with 1M NaOH and then poured into a 50 10 mL of this stock solution was placed into a 100 mL volumetric flask and 0.1M iron chloride was added until the bottom of the meniscus reached the fill line. What was the molarity of absorbing complex in this solution? .0ol33a M mL volumetric flask. Water was added to the fill line to make a stock solution.

Explanation / Answer

6.

A = -log T = - log (I / Io).

We have A = 0.5 and Io = 20 mW/cm2

We need find I = ?

0.5 = -log(I/20 mW/cm2)

I /20 mW/cm2= 0.316

I = 6.32 mW/cm2

7. we have calibration curve and its equation is y=204.5x +0.1, but this is done with 5 cm path length cell.

We have absorbance A, measured with 1cm cell = 0.175

A = e.c.l

0.175 = e.c. 1cm

e.c = 0.175

Suppose, the same solution is measured absorbance using 5 cm cell, then the absorbance would be ;

A = e.c.l

A = 0.175 x 5 cm = 0.875

Therefore, from the calibration curve

Y = 204.5x + 0.1

0.875 = 204.5x + 0.1

X = 0.00379 M

The unknown sample concentration = 3.8 x 10-3 M

8. Acetylsalicylic acid + FeCl3 ----> Violet colored complex

In this reaction, we have the excess of FeCl3, so the limiting reagent is Acetylsalicylic acid.

The stock solution concentration = [Stock] = (wt in g/mol.wt)(1/volume in L)

                                                                                = (0.12 g/180.157 g/mol)(1/0.05 L)

                                                                                = 0.01332 M

Again 10mL of this stock solution is diluted to 100 mL. So the net concentration of acetylsalicylic acid is

                                                                   = 10 mL x 0.01332 M/100 mL = 0.001332 M

Hence, the concentration of the dye = 0.001332 M

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