How many 20 mL extractions of organic solvent are needed to extract at least 99.5% of 50.0 g Compound A from 100 mL of water if K

5 extractions to extract at least 99.5% of 50.0 g Compound A from 100 mL of water

Explanation:

If K=10...

Partition coefficient is defined as the ratio of concentrations of a compound A in two inmiscibles solvents:

K = 10 = Conc. Organic solvent / Conc. Water

Usually organic phase over aqueous phase.

In the first 20mL extraction, the organic solvent will extract:

10 = X / 20mL / (50.0g - X) / 100mL

10 = 100X / (1000-20X)

10000 - 200X = 100X

10000 = 300X

X = 33.3g of compound A are extracted in the first extraction

Remember you want to extract 99.5%, that is 50.0g*99.5% = 49.75g

In the aqueous phase remain: 50-33.3g = 16.7g:

Second extraction:

10 = X / 20mL / (16.7g - X) / 100mL

10 = 100X / (334-20X)

3340 - 200X = 100X

3340 = 300X

11.1g are extracted and will remain: 16.7g - 11.1g = 5.6g

Third extraction:

10 = X / 20mL / (5.6g - X) / 100mL

10 = 100X / (112-20X)

1120 - 200X = 100X

1120 = 300X

3.8g are extracted and will remain: 5.6g - 3.8g = 1.8g

Fourth extraction:

10 = X / 20mL / (1.8g - X) / 100mL

10 = 100X / (36-20X)

360 - 200X = 100X

360 = 300X

1.2g are extracted and will remain: 1.8g -1.2g =  0.6g

Fifth extraction:

10 = X / 20mL / (0.6g - X) / 100mL

10 = 100X / (12-20X)

120 - 200X = 100X

120 = 300X

0.4g are extracted. The total extractions gives:

33.3g + 11.1g + 3.8g + 1.2g + 0.4g = 49.8g

That means, you need to do:

5 extractions to extract at least 99.5% of 50.0 g Compound A from 100 mL of water

They can measure  the amount of energy gained or lost

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.