- #1

- 14

- 0

- Thread starter chiefy
- Start date

- #1

- 14

- 0

- #2

- 14

- 0

any ideas? anyone?

- #3

GCT

Science Advisor

Homework Helper

- 1,728

- 0

Suggestion: For each extraction let x equal the weight extracted into the ether layer. In case (a) the concentration in the ether layer is x/100, and in the water layer is (30-x)/300; the ratio of the quantities is equal to k = 20/13.3.

- #4

- 14

- 0

Well, is this right?

1) k = 20/13.3 = 1.5

x/30-x = .50

x = .5(30-x)

1.5x = 15

x = 10g in ether layer

30 – x = 20g in water layer

x = .5(20 – x)

x = 10 -.5x

1.5x = 10

x = 6.67

20 – x = 13.33

x/13.33 – x =.50

x = .50(13.33 – x)

x = 6.67 - .5x

x = 4.44

13.33 – 4.44 = 8.89

a) 21.1g = three 100-mL portions of ether

x/30-x =1.5

x=1.5(30-x)

x =45-1.5x

2.5x/2.5 =45/2.5

x = 18.0

b) 18.0 g = one 300-mL extraction of ether

- #5

GCT

Science Advisor

Homework Helper

- 1,728

- 0

yep everything seems correct

- Last Post

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 10

- Views
- 3K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 6

- Views
- 4K

- Last Post

- Replies
- 2

- Views
- 20K

- Last Post

- Replies
- 4

- Views
- 6K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K