A nurse wants to add water to 30 ounces of a 10% solution of benzalkonium chloride to dilute it to an 8% soultion. How much water nust she add?(hint water is 0% benzalkonium chloride)
To solve this problem, we need to determine how much water the nurse should add to achieve the desired concentration of 8%. Here's how you can calculate it:
Step 1: Understand the problem
The nurse has a 30-ounce solution of benzalkonium chloride, which is currently at a concentration of 10%. The goal is to dilute this solution by adding water until it reaches a concentration of 8%.
Step 2: Set up the equation
Let's denote the amount of water to be added as "W" (in ounces). Since water does not contain any benzalkonium chloride (0% concentration), the amount of benzalkonium chloride in the diluted solution will remain the same at 30 ounces.
The equation to represent the concentration is as follows:
(amount of benzalkonium chloride / total volume) * 100% = desired concentration
In this case, the desired concentration is 8%.
(30 / (30 + W)) * 100% = 8%
Step 3: Solve for W
First, simplify the equation by dividing both sides by 100%:
30 / (30 + W) = 0.08
Next, cross-multiply:
0.08 * (30 + W) = 30
Now, distribute 0.08 to both terms inside the parentheses:
2.4 + 0.08W = 30
Subtract 2.4 from both sides:
0.08W = 27.6
Finally, divide both sides by 0.08 to solve for W:
W = 27.6 / 0.08
W ≈ 345
Thus, the nurse must add approximately 345 ounces of water to achieve an 8% solution.
Benzalk:
.08(W+30)=.10*30
that says 8percent of the total solution has the same amount of benzyl as did the original solution.
Solve for W.