Question

5. Consider a 0.015 L aqueous solution that contains 1.4 g of caffeine. The solubility of caffeine in water is 2.2 g/100 mL and the solubility in methylene chloride is 10.2 g/100 mL a) How much caffeine can be extracted with 15 mL of methylene chloride? b) How much caffeine will be extracted if the aqueous solution is extracted twice with 7.5 mL of methylene chloride? o) Which is more efective a or b? 2017 Wal-Mart Stores, Inc Neon Pink 635 Distributed by Distribuido po Wal-Mart Stores, Inc Bentonville, AR 7216 8735 ADE IN CHINA MECHO ENCH Walmart çom

Answer 1

Q5

Calculate K coefficient

K = Solubility in methylene chloride/ Solubility in water

K = (10.2/100) /(2.2/100) = 4.636

Now..

a)

n =1

V = 15 mL of methylene chloride

let x be the mass extracted

K = (x)/(15) / (1.4-x) / (15)

4.636 = x/ (1.4-x)

4.636 *1.4 - 4.636 x = x

5.636 x = 4.636 *1.4

x = 5.636 *1.4/6.636

x = 1.18902 g will be extracted

b)

extracted if:

n = 2

V = 7.5 mL of methylene chloride

let x be the mass extracted

K = (x)/(7.5) / (1.4-x) / (15)

4.636 = 2x/ (1.4-x)

4.636 *1.4 - 4.636 x = 2x

6.636 x = 4.636 *1.4

x = 4.636 *1.4/6.636

x = 0.97805

caffeine left = 1.4-0.97805 = 0.4220

This is once, then

repeat

K = (y)/(7.5) / (0.4220-y) / (15)

4.636 = 2y/ (0.4220-y)

4.636 *0.4220- 4.636y = 2y

6.636y = 4.636 *0.4220

y = (4.636 *0.4220) / 6.636

y = 0.2948

Total extracted = x+y =  0.97805+0.2948 = 1.27285 g

clearly, it is better to do 2 extractions