2) The distribution coefficient between methylene chloride and water for solute

Question

2) The distribution coefficient between methylene chloride and water for solute Y is 9. An amount of 84.0 g of Y is dissolved in 160 mL of water.

a) What w eight of Y would be removed from water with a single extraction with 160 - mL of methylene chloride? Report to 1 decimal place.

b) What weight of Y would be removed from water ( the original solution ) with two successive extractions with 80 - mL portions each of methylene chloride? Report to 1 decimal place.

2) You have 127.0 g of solute Q dissolved in 150 - mL of water. The distribution coefficient between ether and water for solute Q is 7.

What volume of ether would you need to extract 103.0 g of solute Q in one extraction from the original 127.0g dissolved in 150 mL water? Provide answer to closest mL. (Show calculations)

Explanation / Answer

Q2.

a)

C = concentration CH2Cl2 / concentration H2O = 9

m = 84 g of Y is dissolved in 160 mL of water..

single extraction... V = 160 mL used in other specie

so

concentration CH2Cl2 / concentration H2O = 9

(y / 160) / ((84-y) / (160)) = 9

y / (84-y) = 9

y = 84*9 - 9y

10y = 84*9

y = 84*9/10 = 75.6 g in organic layer, (84-75.6) = 8.4 g in aqeuous layer

b)

for n = 2 ?

initially:

concentration CH2Cl2 / concentration H2O = 9

(y)/80 / (80-y) / 160 = 9

y / (80-y) = 9*2

y = 18*80 - 18y

19y = 18*80

y = 18/19*80 = 75.789 g in organic layer

84-75.789 = 8.211 in aqeuous layer

now..

second dilution

(z)/80 / (8.211-z)/160 = 9

z/(8.211 -z) = 80/160*9

z =  80/160*9*8.211 - 80/160*9*z

(1+ 80/160*9)z = 80/160*9*8.211

z = 80/160*9*8.211 / ((1+ 80/160*9)) = 6.718 g in organic layer

so...

water left = 8.211 -6.718 = 1.493 g in water

so total removed = 84-1.493 = 82.507

Q2.

m = 127 g of Q in V = 150 mL water

Q = 7

fin dvolume of ther is Q = 103 g of Q is required...

concentration CH2Cl2 / concentration H2O = 7

103 / (volume organic) / (127-103) / 150) = 7

solv efor volume

volume organic = 103 / (7*( (127-103) / 150) ))

V = 91.964mL required

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