Alcohol and Water Mixing Problem

Let's analyze the scenario step by step to determine the final amounts of alcohol and water in the containers.

Initial State

• Container A: Contains 1 L of alcohol. This is equal to 1000 mL of alcohol and 0 mL of water.
• Container B: Contains 1 L of water. This is equal to 0 mL of alcohol and 1000 mL of water.

First Transfer (A to B)

100 mL of alcohol is transferred from container A to container B.

• After transfer, Container A has $1000 \text{ mL} - 100 \text{ mL} = 900 \text{ mL}$ alcohol and 0 mL water.
• Container B now contains its initial 1000 mL water plus the 100 mL alcohol transferred from A.
• Total volume in Container B is $1000 \text{ mL} + 100 \text{ mL} = 1100 \text{ mL}$.
• The mixture in Container B consists of 100 mL alcohol and 1000 mL water.

Mixture Composition in Container B

Container B is mixed well. The concentration of alcohol and water in the mixture is:

• Concentration of alcohol in B = $\frac{\text{Volume of Alcohol}}{\text{Total Volume}} = \frac{100 \text{ mL}}{1100 \text{ mL}} = \frac{1}{11}$
• Concentration of water in B = $\frac{\text{Volume of Water}}{\text{Total Volume}} = \frac{1000 \text{ mL}}{1100 \text{ mL}} = \frac{10}{11}$

Second Transfer (B to A)

100 mL of the mixture is transferred back from container B to container A.

• The amount of alcohol in the 100 mL mixture transferred from B to A is $100 \text{ mL} \times \text{Concentration of alcohol in B} = 100 \text{ mL} \times \frac{1}{11} = \frac{100}{11} \text{ mL}$.
• The amount of water in the 100 mL mixture transferred from B to A is $100 \text{ mL} \times \text{Concentration of water in B} = 100 \text{ mL} \times \frac{10}{11} = \frac{1000}{11} \text{ mL}$.

Final State

Let's determine the final amounts of alcohol and water in both containers after the second transfer.

• Container A:
  • Initial alcohol (after first transfer): 900 mL
  • Alcohol transferred from B: $\frac{100}{11}$ mL
  • Final alcohol in A = $900 + \frac{100}{11} = \frac{9900 + 100}{11} = \frac{10000}{11} \text{ mL}$.
  • Initial water (after first transfer): 0 mL
  • Water transferred from B: $\frac{1000}{11}$ mL
  • Final water in A = $0 + \frac{1000}{11} = \frac{1000}{11} \text{ mL}$.
• Container B:
  • Initial alcohol (after first transfer): 100 mL
  • Alcohol transferred to A: $\frac{100}{11}$ mL
  • Final alcohol in B = $100 - \frac{100}{11} = \frac{1100 - 100}{11} = \frac{1000}{11} \text{ mL}$.
  • Initial water (after first transfer): 1000 mL
  • Water transferred to A: $\frac{1000}{11}$ mL
  • Final water in B = $1000 - \frac{1000}{11} = \frac{11000 - 1000}{11} = \frac{10000}{11} \text{ mL}$.

Comparison

We need to compare the amount of alcohol in container B with the amount of water in container A.

• Amount of alcohol in container B = $\frac{1000}{11} \text{ mL}$.
• Amount of water in container A = $\frac{1000}{11} \text{ mL}$.

The amount of alcohol in container B is $\frac{1000}{11}$ mL, and the amount of water in container A is also $\frac{1000}{11}$ mL. These amounts are the same.

Summary of Amounts

The amount of alcohol in container B is equal to the amount of water in container A.