Question

A chemist has 5 l of a solution that contains 0.25 l of acetic acid. the chemist needs to prepare 9 l of a solution with the same percent concentration of acetic acid. how much acidic acid must the 9 l of the solution contain

Answer 1

Answer:- 0.45L

Solution:- From information, 0.25 L of acetic acid are present in 5 L of a solution. We could calculate the percentage of acetic acid in this solution using the formula:

percent by volume =$(\frac{acid volume}{solution volume})100$

Let's plug in the values in it:

percent by volume =$(\frac{0.25}{5})100$

= 5%

Now, it asks to calculate the volume of acetic acid present in 9 L solution with the same percentage. Let's use the same formula again and plug in the values in it:

$5=(\frac{acid volume}{9})100$

acid volume = $\frac{5*9}{100}$

acid volume = 0.45 L

So, 9 L of the solution must contain 0.45 L of acetic acid.