An explanation with your calculations so I can go through it to understand pleas

Question

An explanation with your calculations so I can go through it to understand please and thank you!!! Questions 1. Suppose that 9.75 grams of a substance A is dissolved in 100 mL of water and extracted once with 90 mL of diethyl ether. Assuming K-3.25 favoring ether, how much of Ais extracted into the ether? Show calculation. Answer 2. What weight of A would be extracted from the original aqueous solution (above) if extracted with 3 x 30 mL of ether? Show calculation. Show the amount extracted in each extraction. lst. 2nd 3rd grams Sum (total extracted) 3. What general conclusions can you draw concerning extraction from the answers to question and (2) above?

Explanation / Answer

1) We have 9.75 g of substance A dissolved in 100 mL water; therefore, concentration of A = 9.75 g/100 mL = 0.0975 g/mL.

Let us consider the partition as

A (aq) <=====> A (organic)

Let C1 be the concentration in the aqueous phase after 90 mL diethyl ether is added. Let x g of solute remain in the aqueous phase after extraction so that

C1 = x g/100 mL = x/100 g/mL.

The amount of substance A partitioned into the organic layer is (9.75 – x). Let C2 be the concentration in the organic layer such that

C2 = (9.75 – x) g/90 mL = (9.75 – x)/90 g/mL.

The partition co-efficient is

K = A(organic)/A(aqueous) = 3.25

===> 3.25 = C2/C1

===> 3.25 = [(9.75 – x)/90]/(x/100)

===> 3.25 = 100*(9.75 – x)/90x

===> 3.25*90*x = 100*(9.75 – x)

===> 292.5x = 975 – 100x

===> 392.5 x = 975

===> x = 975/392.5 = 2.484

The mass of A remaining in the aqueous layer after a single extraction = 2.484 g and the mass extracted into the ether layer = (9.75 – 2.484) g = 7.266 g (ans).

The efficiency of the extraction = (7.266/9.75)*100% = 74.523% (ans).

2) We need to start like above; here the volume of diethyl ether used will be 3*30 mL.

Work out step-by-step:

Step 1: Original mass = 9.75 g; volume of water = 100 mL and volume of ether = 30 mL.

Hence, C1 = x/100 (as above), but C2 = (9.75 – x)/30

Therefore, 3.25 = [(9.75 – x)/30]/(x/100)

===> 3.25*30*x = 100*(9.75 – x)

===> 97.5x = 975 – 100x

===> 197.5x = 975

===> x = 975/197.5 = 4.9367 4.937

Mass extracted = (9.75 – 4.937) g = 4.813 g (ans).

Mass remaining in aqueous layer = 4.937 g

Step 2: Mass in aqueous layer = 4.813 g; volume of aqueous layer = 100 mL and volume of ether = 30 mL.

C1 = x1/100 and C2 = (4.937 – x1)/30

Again, 3.25 = [(4.937 – x1)/30]/(x1/100)

===> 3.25*30*x1 = 100*(4.937 – x1)

===> 97.5x1 = 493.7 – 100x1

===> 197.5x1 = 493.7

===> x1= 493.7/197.5 = 2.4997 2.500

Mass remaining in aqueous layer = 2.500 g.

Mass extracted in ether layer = (4.937 – 2.500) = 2.437 g (ans).

Step 3: C1 = x2/100 and C2 = (2.437 – x2)/30

Therefore, 3.25*30*x2 = 100*(2.437 – x2)

====> 97.5x2 = 243.7 – 100x2

====> x2 = 243.7/197.5 = 1.245

Mass remaining = 1.245 g

Mass extracted = (2.500 – 1.245) g = 1.255 g (ans).

Total mass extracted = (4.813 + 2.437 + 1.255) g = 8.505 g (ans).

Efficiency of extraction = (8.505/9.75)*100 = 87.23% (ans).

3) Multiple extractions with small volume of organic solvent leads to better extraction of the substance into the organic layer. In problem 2, we used 3*30 mL of solvent and extracted 87.23% solute while we extracted only 74.523% solute in a single extraction with 90 mL solvent (as in problem 1 above).

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