Question

Substance s has Imax = 485 nm. Two 50 mL aliquots of an aqueous solutions are transferred to two different 100 mL volumetric flasks. 25.0 mL of 200.0 ppm S are added to the second flask. Both flasks are then diluted to 100.0 mL with distilled water. At 485 nm, the first flask has A = 0.589 and the second has A = 3.72. Calculate the concentration of Sin the original solution. Select one: O A. None of these O B. 176 ppm O C. 18.8 ppm O D. 6.81 ppm O E. 88.3 ppm

Answer 1

We assume that ppm is given in terms of volume because the problem deals with concentration and volume .

Beer Lamberts law is used for this question, A $\alpha$ C

where , A is absorptivity and C is concentration. We introduce a constant k to equation to change proportionality to equality

A = k C

Now lets look into our problem, consider y to be ppm of the original solution. y ppm of S means that 1ml of solution contains is y $\times$ 10^-6mL of S.

Now we have two flasks,

In the first flask, there is 50 ml of y ppm solution .It contains 50$\times$y mL of S in it.Next 50 mL of distilled water is added to make up to 100 mL.Finally ,there is 50$\times$y mL of S in 100 mL of solution .

So final concentration is $\frac{(50 \times y) \: mL \, of \, S }{100 mL \, of \,solution}$

In the second flask, there is 50 ml of y ppm solution .It contains (50$\times$y) x 10^-6 mL of S in it.Next 25 mL of 200 ppm solution is added. This 25 mL contains (200 x 25) x 10^-6 mL of S = 5000 x 10^-6 mL of S.So total S adds up to (50$\times$y) x 10^-6 mL + x 10^-6 mL = (50$\times$y+5000) x 10^-6 mL .Then volume is made up to 100 mL by adding distilled water.

So final concentration is $\frac{(50 \times y + 5000) \: mL \, of \, S }{100 mL \, of \,solution}$

Now use A = k C

For first flack, A= 0.589 A=kC 0.589 = kx 50 g -0 100 For second Plask, A= 3.72 3.72= kx 50 g + 5000 100 eliminate k To and find JI divide equations land 2, 0.589 K x 50g 0. (2) 3.72 K & 50y + 5000 co0 Solve for YI 18.81 ppm ya option C is correct

y is the orginal concentration and y is obtained as 18.8. Answer is option C.

Hope ir helped you and if it does,Plz give like

All the Best and Thank You