1. The distribution coefficient, K, is 10 for compound Y. What weight of compound Y would be removed from a solution of 4.0 g of Y in 100 mL of water by a single extraction with 100 mL of ether?
2. Reconsider question 1 above; what weight of compound would be removed by two (double) extractions using 50 mL of ether each time?
Help! Organic chemistry extraction!!?
2016-06-15 4:06 am
回答 (1)
2016-06-15 4:22 am
1.
Let x g be the weight of compound Y removed.
K = (x/100) / [(4.0 - x)/100] = 10
x/(4.0 - x) = 10
x = 40 - 10x
11x = 40
x = 3.64
Weight of compound Y removed = 3.84
====
2.
Let y g and z g be the weight of compound Y removed in the first and second extraction respectively.
First extraction :
K = (y/50) / [(4.0 - y)/100] = 10
y/(4.0 - y) = 5
y = 20 - 5y
6y = 20
y = 3.33
Weight of compound Y extracted in the second extraction
= (4.0 - 3.33) × (3.33/4.0) g
= 0.56 g
Total weight of compound Y extracted = (3.33 + 0.56) g = 3.89 g
Let x g be the weight of compound Y removed.
K = (x/100) / [(4.0 - x)/100] = 10
x/(4.0 - x) = 10
x = 40 - 10x
11x = 40
x = 3.64
Weight of compound Y removed = 3.84
====
2.
Let y g and z g be the weight of compound Y removed in the first and second extraction respectively.
First extraction :
K = (y/50) / [(4.0 - y)/100] = 10
y/(4.0 - y) = 5
y = 20 - 5y
6y = 20
y = 3.33
Weight of compound Y extracted in the second extraction
= (4.0 - 3.33) × (3.33/4.0) g
= 0.56 g
Total weight of compound Y extracted = (3.33 + 0.56) g = 3.89 g
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