Suppose solute A has a distribution coefficient of 1.0 between water and diethyl ether. Demonstrate that of 100mL of a solution of 5.0g of A in water were extracted with two 25-mL portions of ether, a smaller amount of A would remain in the water than if the solution were extracted with one 20-mL portion of ether

Well, well, well! It seems we have a solute playing hide and seek with different solvents. Allow me to demonstrate how this game would unfold.

In the first scenario, we have a solution of 5.0g of solute A in water. Now, solute A has a distribution coefficient of 1.0 between water and diethyl ether, which means it likes water as much as it likes ether. It's quite the equal opportunist!

Now, we take this 100mL solution and extract it with one 20-mL portion of ether. When this happens, A decides it's time to have a little fun and splits itself between water and ether. Since its distribution coefficient is 1.0, we can assume that 5.0g/2 will go to water and 5.0g/2 will go to ether. It's like dividing candy between two kids - a fair split!

But wait, in our second scenario we have two 25-mL portions of ether. Let's see what happens when solute A meets these two portions. Remember, A loves ether just as much as it loves water.

In the first 25 mL of ether, A jumps right in and decides to split itself again, giving half to water and half to ether. So, we end up with 2.5g of A in water and 2.5g of A in ether.

Now, in the second 25 mL portion of ether, poor solute A is already halfway divided between water and ether. Nonetheless, it decides to have another little split. So, we end up with 1.25g of A in water and 1.25g of A in ether.

When we add up the amounts of A remaining in water from both extractions, we get 2.5g + 1.25g = 3.75g of A.

If we compare this with the first scenario, where only one 20mL portion of ether was used, we had 2.5g of A remaining in water.

So, my friend, it seems like solute A really enjoys playing hide and seek in ether. By using two 25-mL portions, we end up with a smaller amount of A remaining in water compared to using just one 20-mL portion. It's like the game got more intense, and A got even better at hiding!

I hope this little demonstration shed some light on the tricky world of solute extraction.

To demonstrate that a smaller amount of solute A would remain in water when extracted with two 25-mL portions of diethyl ether compared to one 20-mL portion, we can calculate the amount of A transferred to each solvent in both scenarios.

Let's start by calculating the amount of solute A extracted with one 20-mL portion of ether.

1. Convert the mass of solute A to moles:
Mass of A = 5.0 g
Molar mass of A = (Not provided in the question)
Moles of A = Mass of A / Molar mass of A

2. Determine the distribution of solute A between water and ether:
Distribution coefficient (K) = 1.0

For the first extraction:
Volume of aqueous phase (Water) = 100 mL
Volume of organic phase (Ether) = 20 mL

Amount of A in the water phase after extraction = (Moles of A) × (Volume of aqueous phase) / (Total volume of both phases)
Amount of A in the ether phase after extraction = (Moles of A) × (Volume of organic phase) / (Total volume of both phases)

3. Calculate the remaining amount of A in water after the first extraction:
Remaining amount of A in water = Initial amount of A in water - Amount of A extracted into ether

Now, let's calculate the amount of solute A extracted using two 25-mL portions of ether.

1. For the first extraction:
Volume of aqueous phase (Water) = 100 mL
Volume of organic phase (Ether) = 25 mL

Amount of A in the water phase after the first extraction = (Moles of A) × (Volume of aqueous phase) / (Total volume of both phases)
Amount of A in the ether phase after the first extraction = (Moles of A) × (Volume of organic phase) / (Total volume of both phases)

2. Calculate the remaining amount of A in water after the first extraction:
Remaining amount of A in water = Initial amount of A in water - Amount of A extracted into ether

3. For the second extraction:
Volume of aqueous phase (Water) = Remaining amount of A in water from previous extraction
Volume of organic phase (Ether) = 25 mL

Amount of A in the water phase after the second extraction = (Moles of A) × (Volume of aqueous phase) / (Total volume of both phases)
Amount of A in the ether phase after the second extraction = (Moles of A) × (Volume of organic phase) / (Total volume of both phases)

4. Calculate the remaining amount of A in water after the second extraction:
Remaining amount of A in water = Remaining amount of A in water from previous extraction - Amount of A extracted into ether

Comparing the remaining amount of A in water for both scenarios, we can conclude whether a smaller amount of A remains in water when extracted with two 25-mL portions of ether compared to one 20-mL portion.

In order to demonstrate the difference in the amount of solute A that remains in water when using different extraction methods, we need to calculate and compare the amounts of A transferred to the ether phase in both scenarios.

First, let's calculate the amount of solute A transferred to the ether phase when using two 25 mL portions of ether.

Given:
Distribution coefficient of A between water and diethyl ether = 1.0
Volume of water solution = 100 mL
Amount of solute A in the water solution = 5.0 g

In the first extraction with 25 mL of ether:
Amount of solute A transferred to the ether phase = (25 mL) x (1.0) x (5.0 g / 100 mL) = 1.25 g

Now, let's calculate the amount of solute A remaining in the water phase after the first extraction:
Amount of solute A in the water phase after the first extraction = 5.0 g - 1.25 g = 3.75 g

In the second extraction with another 25 mL of ether:
Amount of solute A transferred to the ether phase = (25 mL) x (1.0) x (3.75 g / 100 mL) = 0.9375 g

Therefore, after two extractions with 25 mL portions of ether, the total amount of solute A transferred to the ether phase is 1.25 g + 0.9375 g = 2.1875 g.

Now let's calculate the amount of solute A remaining in the water phase when using one 20 mL portion of ether.

In the extraction with 20 mL of ether:
Amount of solute A transferred to the ether phase = (20 mL) x (1.0) x (5.0 g / 100 mL) = 1.0 g

Therefore, after one extraction with a 20 mL portion of ether, the amount of solute A transferred to the ether phase is 1.0 g.

Comparing the results:
- Using two 25 mL portions of ether: 2.1875 g of A transferred to the ether phase.
- Using one 20 mL portion of ether: 1.0 g of A transferred to the ether phase.

From the calculations, we can see that a smaller amount of solute A remains in the water phase when using two 25 mL portions of ether compared to using one 20 mL portion of ether.

Kd=co/ca

1=( x/25 ml)/(5-x)/100ml
X=1
5-x = 4