Question

An aqueous solution containing 5.0 g of solute in 100 ml is extracted with three 25 ml portion of diethyl ether. what is the total amount of solute that will be extracted by the ether, k = 1.0 ?

Answer 1

Answer:

The correct answer is 2.44 grams.

Explanation:

The partition coefficient, K = concentration of solute in ether / concentration of solute in water

As partition coefficient is 1, therefore, the concentration of solute in both the solvents would be similar.  

Thus, when 25 milliliters of ether is added to 100 ml of water Wether/25 = Wwater/100

However, Wwater + Wether = 5.00 g

Wwater = 5.00 g - Wether

So, Wether/25 = 5.00 - Wether/100

Wether = 1.00 g

Thus, Wwater = 5.00 - 1.00 = 4.00 grams

So, for the first time, the solute extracted by ether is 1.0 gram

Now add 25 milliliters of ether to 4.00 grams of solute of 100 ml water,

Wwater + Wether = 4.00 g

Wwater = 4.00 g - Wwater

So, Wether/25 = 4.00 - Wether/100

Wether = 0.8

Thus, Wwater = 4.0 - 0.8 = 3.2 grams

So, for the second time the amount of solute extracted by ether is 0.8 gram.  

Now, add 25 ml of ether to 3.2 grams of solute of 100 ml water

Wwater + Wether = 3.20 g

Wwater = 3.20 - Wether

So, Wether/25 = 3.20 - Wether/100

Thus, Wwater = 3.20 - 0.64 = 2.56 grams

So, for the third time, the amount of solute extracted by ether is 0.64 grams.  

Therefore, total weight of solute extracted by ether is 1.00 + 0.80 + 0.64 = 2.44 grams.  

Answer 2

First, in this case, we define the K constant as the solubility of the solute in water divided by the solubility of the solute in ether.

K = (X grams of solute / 75 mL of ethyl ether) / (5 g of solute / 100 mL of water)

K = (X / 75) / (5 / 100)

1 = (X / 75) / (5 / 100)

5 / 100 = X / 75

0,05 = X /75

X = 0,05 × 75 = 3.75 g of solute that will be extracted by the ether