Question

Ruth has a beaker containing a solution of $800$ mL of acid and $200$ mL of water. She thinks the solution is a little strong, so she drains $300$ mL from the beaker, adds $300$ mL of water, and stirs the solution. Ruth thinks the solution is still too strong, so again she drains $400$ mL from the beaker, and adds $400$ mL of water. Ruth again thinks the solution is still too strong, so again she drains $100$ mL from the beaker, and adds $100$ mL of water. How many mL of water are now in the beaker?

Answer 1

To solve the problem, we will track the volumes and concentrations of acid and water in the beaker after each operation.

Initially, Ruth has:

• 800 mL of acid
• 200 mL of water

The total initial volume of the solution is: \[ 800 \text{ mL (acid)} + 200 \text{ mL (water)} = 1000 \text{ mL (total solution)} \]

Step 1: Drain 300 mL and add 300 mL of water

The concentration of the acid in the initial solution is: \[ \text{Concentration of acid} = \frac{800}{1000} = 0.8 \] When 300 mL is drained from the beaker, the amount of acid and water removed is proportional to their concentrations in the solution.

Since \(300 \text{ mL}\) is drained, the amounts of acid and water removed can be calculated as follows:

Acid removed: \[ 800 \text{ mL} \times \frac{300 \text{ mL}}{1000 \text{ mL}} = 240 \text{ mL of acid} \] Water removed: \[ 200 \text{ mL} \times \frac{300 \text{ mL}}{1000 \text{ mL}} = 60 \text{ mL of water} \]

This means after draining 300 mL:

• Remaining acid = \(800 - 240 = 560 \text{ mL}\)
• Remaining water = \(200 - 60 = 140 \text{ mL}\)

Now, she adds 300 mL of water: \[ \text{New water amount} = 140 + 300 = 440 \text{ mL} \]

The new total volume in the beaker is: \[ \text{New total volume} = 560 \text{ mL (acid)} + 440 \text{ mL (water)} = 1000 \text{ mL} \]

Step 2: Drain 400 mL and add 400 mL of water

Now, we again calculate the concentrations: \[ \text{Concentration of acid} = \frac{560}{1000} = 0.56 \] When she drains 400 mL, the amounts of acid and water removed are: Acid removed: \[ 560 \text{ mL} \times \frac{400 \text{ mL}}{1000 \text{ mL}} = 224 \text{ mL of acid} \] Water removed: \[ 440 \text{ mL} \times \frac{400 \text{ mL}}{1000 \text{ mL}} = 176 \text{ mL of water} \]

After draining:

• Remaining acid = \(560 - 224 = 336 \text{ mL}\)
• Remaining water = \(440 - 176 = 264 \text{ mL}\)

Now add 400 mL of water: \[ \text{New water amount} = 264 + 400 = 664 \text{ mL} \]

The new total volume in the beaker is: \[ \text{New total volume} = 336 \text{ mL (acid)} + 664 \text{ mL (water)} = 1000 \text{ mL} \]

Step 3: Drain 100 mL and add 100 mL of water

Now for the final step, we compute the new concentrations: \[ \text{Concentration of acid} = \frac{336}{1000} = 0.336 \] When she drains 100 mL, the amounts of acid and water removed are: Acid removed: \[ 336 \text{ mL} \times \frac{100 \text{ mL}}{1000 \text{ mL}} = 33.6 \text{ mL of acid} \] Water removed: \[ 664 \text{ mL} \times \frac{100 \text{ mL}}{1000 \text{ mL}} = 66.4 \text{ mL of water} \]

After draining:

• Remaining acid = \(336 - 33.6 = 302.4 \text{ mL}\)
• Remaining water = \(664 - 66.4 = 597.6 \text{ mL}\)

Now add 100 mL of water: \[ \text{New water amount} = 597.6 + 100 = 697.6 \text{ mL} \]

Thus, the final answer, the amount of water in the beaker now, is: \[ \boxed{697.6} \text{ mL} \]