Question

You need a 55% alcohol solution. On hand, you have a 605 mL of a 90% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

A) Write an equation using the information as it is given above that can be used to solve this problem. Use as your variable to represent the amount of pure water you need to use.
what is the Equation
You will need how much
mL of pure water
to obtain
mL of the desired 55% solution.

Answer 1

To create a 55% alcohol solution, around 385 mL of pure water must be added to 605 mL of a 90% alcohol mixture, giving a total volume of 990 mL for the desired solution.

Step-by-step explanation:

To answer the question, we need to set up an equation based on the concentration of alcohol in the final solution. If we designate x as the amount of pure water we need to add, here is how we can develop the equation:

The initial volume of alcohol in the 90% solution is 0.90 × 605 mL = 544.5 mL. To decrease the concentration to 55%, we need the total volume of the solution to be such that the alcohol comprises 55% of that total volume.

So the equation for the final concentration (Cfinal) is:

Cfinal = (Volume of alcohol) / (Volume of original solution + Volume of water added)

Substituting the known values, we get:

0.55 = 544.5 mL / (605 mL + x)

To find x, we multiply both sides by (605 mL + x) and then isolate x:

544.5 mL = 0.55 × (605 mL + x)

544.5 mL = 332.75 mL + 0.55x

211.75 = 0.55x

And therefore:

x = 211.75 / 0.55

We can then solve for x to find the amount of pure water needed. Upon solving, we find that:

x ≈ 385 mL

So, about 385 mL of pure water is needed to achieve a 55% alcohol solution.

The total volume of the desired solution can be calculated by adding the volume of the original 90% solution to the volume of water added. That is:

Total Volume = 605 mL + 385 mL

= 990 mL