Question

1) Complete the calculations using the information provided (Show all work) Time Absorbance 0 0.707 1...

1) Complete the calculations using the information provided (Show all work)

Time Absorbance
0 0.707
1 0.705
2 0.708
3 0.702
4 0.698
5 0.699
7 0.699
9 0.673
11 0.671
13 0.661
15 0.650
20 0.658
25 0.655
30 0.640
35 0.643
40 0.623

Absorbance of Cu2+ after unadsorbed copper was collected at end of 40 minutes= 0.537

Absorbance of the initial solutiion = 0.707Mass of CuCl2 * 2H2O used to prepare the 200mL solution= 1.706

(A) Initial concentration of CuCl2 * 2H2O, M (moles/L)

(B) Calculated omlar extinction coefficient of CuCl2 * 2H2O

(C) Absorbance of the solution containing residual CuCl2 * 2H2O

(D) Final concentration of CuCl2 * 2H2O, M (moles/L)

(E) initial mass of Cu2+ in solution (per 200 mL)

(F) Final mass of Cu2+ in solution

(G) Mass of Cu2+ absorbed by the polymer

(H) Mass of super absorbent polymer used (g)

(I) Mass of Cu2+ absorbed per gram of the polymer

0 0
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Answer #1

a) The initial concentration of CuCl2.2H2O:

m2 mmolar M = 1.7069 170.48g/mol 0.2001 -= 0.0500 M = 0. V(L)

b) The molar extinction coefficient can be calculated from the Lambert-Beer Law, assuming l = 1.0 cm (which is its usual value):

A 1.0.1cm 0.707 0.0500M = 14.1 M-cm-1

c) The absorbance of the solution containing residual CuCl2 was 0.537

d) The final concentration can be calculated from the Lambert-Beer law, using the final concentration:

A 0.537 C=4 -- 0.537 2. lcm .14.1 M-1m-1 = 0.0380 M

e) The initial mass of Cu(II) in solution can be calculated taking into account that there are 63.55 grams of Cu in 170.48 g of CuCl2.H2O, which means that in 1.706 g of the compound there are:

63.559 m = 1.7069 170.489 = 0.636 g

f) For the final mass, we can determine the mass of Cu that would be present in the final solution, from its concentration (which, in molar terms, is the same as the concentration of CuCl2):

m = M.V(L).mmolar = 0.0380 -.0.200L - 63.55_9, = 0.483 g mol

g) The mass absorbed by the polymer is the difference between the initial and final mass: 0.636 g - 0.483 g = 0.153 g

Items h and i require the mass of polymer used, which is not given.

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