# a bottle contains 1000ml of pure ethanol. 300 ml is removed and the bottle is topped up with pure water. the mixture is stirred.

what is the volume of ethanol in the bottle if the process is repeated five times?

how many times should the process be repeated for there to be less than 1 ml of ethanol in the bottle?

## first procedure:

amount alcohol removed = 300 ml

amount of alcohol left = 700 ml or .7(1000)

topped up with 300 ml of water

alcohol left in the 1000 ml = 700 ml

percentage of alcohol in the mixture = 70% = .7

2nd procedure:

amount of alcohol removed = .7(300) ml

amount of alcohol left = 700 - .7(300) = 100(10 - .7(3))

=490

topped up with 300 ml of water

alcohol left in the 1000 ml = 490 ml

percentage of alcohol in the mixture = 490/1000 = .49 = .7^2

3rd procedure:

amount of alcohol removed = .49(300)

amount of alcohol left = 490 = .49(300) = 343

topped up with 300 of water

alcohol left in the 1000 ml mixture = 343

percentage of alcohol in the mixture = 343/1000 = .343 = .7^3

.....

percentage of alcohol in the mixture after the 5th procedure = .7^5 or .16807

volume left = .16807(1000) = 168.07 ml

we want .7^n (1000) < 1

.7^n ≤ .001

let's look at .7^n = .001

take logs of both sides and use log rules:

n log.7 = log .001

n = log .001/log .7

n = 19.3

so to be less than 1 ml you would have to repeat 20 times.

## Thank you!

## To determine the volume of ethanol in the bottle after repeating the process five times, we need to calculate the amount of ethanol remaining after each step.

Let's denote the initial volume of ethanol as V₀ = 1000 ml.

After removing 300 ml and replacing it with water, the volume of ethanol remaining, V₁, can be calculated using the following equation:

V₁ = V₀ - 300 ml

V₁ = 1000 ml - 300 ml

V₁ = 700 ml

After repeating this process four more times, we can calculate the volume of ethanol after each step:

V₂ = V₁ - 300 ml

V₃ = V₂ - 300 ml

V₄ = V₃ - 300 ml

V₅ = V₄ - 300 ml

Now, let's substitute the values into the equation:

V₂ = 700 ml - 300 ml

V₂ = 400 ml

V₃ = 400 ml - 300 ml

V₃ = 100 ml

V₄ = 100 ml - 300 ml

V₄ = -200 ml

V₅ = -200 ml - 300 ml

V₅ = -500 ml

The volume of ethanol cannot be negative, so after the fifth repetition, there is no ethanol left in the bottle.

To find out how many times the process should be repeated for there to be less than 1 ml of ethanol in the bottle, we can set up an equation using the initial volume of ethanol:

V = 1000 ml - 300x ml

Where V represents the volume of ethanol in the bottle after x repetitions.

We want to find the value of x when V < 1 ml:

1000 ml - 300x < 1 ml

Subtracting 1000 ml from both sides:

-300x < -999 ml

Dividing both sides by -300 (and reversing the inequality):

x > 999 ml / 300 ml

x > 3.33

Therefore, the process should be repeated at least four times (x > 3.33) for there to be less than 1 ml of ethanol in the bottle.

## To solve these questions, we need to understand how the volume of ethanol changes after each repetition.

Let's break down the first question:

1. Initial volume of ethanol = 1000 ml

2. 300 ml of ethanol is removed, leaving 1000 ml - 300 ml = 700 ml of ethanol in the bottle.

3. The bottle is then topped up with pure water, which adds 300 ml of water to the mixture.

4. After stirring, the total volume of the mixture remains constant at 1000 ml.

5. Therefore, the volume of ethanol in the mixture is still 700 ml after the first repetition.

Now, let's determine the volume of ethanol after five repetitions:

1st repetition: 700 ml

2nd repetition: 70% of 700 ml = 490 ml (70% of the previous volume)

3rd repetition: 70% of 490 ml = 343 ml (70% of the previous volume)

4th repetition: 70% of 343 ml = 240 ml (70% of the previous volume)

5th repetition: 70% of 240 ml = 168 ml (70% of the previous volume)

After five repetitions, there will be approximately 168 ml of ethanol in the bottle.

Now, let's move on to the second question:

We want to find out how many repetitions are needed for there to be less than 1 ml of ethanol in the bottle.

Using the formula mentioned earlier (70% of the previous volume), we can calculate the volume of ethanol after each repetition and track it until it becomes less than 1 ml:

1st repetition: 700 ml

2nd repetition: 490 ml

3rd repetition: 343 ml

4th repetition: 240 ml

5th repetition: 168 ml

6th repetition: 117.6 ml

7th repetition: 82.32 ml

8th repetition: 57.624 ml

9th repetition: 40.3368 ml

10th repetition: 28.23576 ml

11th repetition: 19.765032 ml

12th repetition: 13.835522 ml

13th repetition: 9.6848654 ml

14th repetition: 6.7784058 ml

15th repetition: 4.74488406 ml

16th repetition: 3.321418842 ml

17th repetition: 2.3249931894 ml

18th repetition: 1.6274952326 ml

19th repetition: 1.13924666282 ml

20th repetition: 0.797472663974 ml

21st repetition: 0.558230864782 ml

Therefore, after 21 repetitions, there will be less than 1 ml of ethanol in the bottle.

## Oh, I see what you're trying to do here. You want to play a little game of hide and seek with the ethanol, huh? Alright, challenge accepted!

Let's break it down step by step and see how much ethanol remains after each round of dilution:

Round 1:

After removing 300 ml of pure ethanol, we are left with 1000 ml - 300 ml = 700 ml of ethanol.

Round 2:

Again, we remove 300 ml, leaving us with 700 ml - 300 ml = 400 ml of ethanol.

Round 3:

Yet another 300 ml is tossed out, leaving 400 ml - 300 ml = 100 ml of ethanol.

Round 4:

Once again, we bid farewell to 300 ml, resulting in 100 ml - 300 ml = -200 ml of ethanol. But hold on a second, negative volume? That's impossible! Our ethanol has vanished into thin air!

So, to answer your first question, after five rounds of dilution, there is no ethanol left in the bottle. It has completely disappeared!

Now, let's move on to your second question. To have less than 1 ml of ethanol, we need to continue the process until we reach that point. Let's do some calculations:

Round 1: 1000 ml - 300 ml = 700 ml of ethanol.

Round 2: 700 ml - 300 ml = 400 ml of ethanol.

Round 3: 400 ml - 300 ml = 100 ml of ethanol.

Round 4: 100 ml - 300 ml = -200 ml of ethanol. Ethanol gone!

So, it looks like we would need to repeat the process four times for there to be less than 1 ml of ethanol in the bottle. But remember, negative volumes are prohibited in this universe, so there's no way to have less than 1 ml through dilution alone. Ethanol has a knack for elusiveness!

Hope that brings some laughter to your scientific endeavors!