100mL of a solution contains 30 mL of acid; this is called a 30% acid solution because
part 30
---- = ---- = 0.30.
whole 100
How many milliliters of acid x would need to be added to the solution to turn it into a 50% solution ? (HInt: Adding acid increases both the volume of acid and the volume of the complete solution.)
didn't I answer this?
To solve this problem, we can use the concept of proportions.
Let's assume that x milliliters of acid need to be added to the solution.
According to the given information, the initial solution is a 30% acid solution, which means that 30 mL of acid is present in a 100 mL solution.
When x mL of acid is added to the solution, the total volume of the solution will be 100 mL (initial solution) + x mL (added acid).
The new solution will be a 50% acid solution, which means that 50 mL of acid should be present in the new solution.
Using the proportion:
30 mL / 100 mL = x mL / (100 mL + x mL)
We can cross-multiply to solve for x:
30 mL * (100 mL + x mL) = 100 mL * x mL
3000 mL + 30x = 100x
Subtract 30x from both sides:
3000 mL = 70x
Divide both sides by 70:
3000 mL / 70 = x
x ≈ 42.857 mL
Therefore, approximately 42.857 milliliters of acid would need to be added to the solution to turn it into a 50% solution.