Chemist has 60 g of solution that is 70% acid.
How much water should be added to make a solution that is 40% acid?
To solve this problem, we can set up an equation based on the amount of acid in the initial solution and the final desired concentration.
Let's denote the amount of water to be added as x grams.
The amount of acid in the initial solution is equal to 70% of 60 g, or 0.7 * 60 = 42 g.
In the final solution, the total amount of solution will be 60 g + x g, and the amount of acid will be 40% of this total amount.
So, 40% of (60 g + x g) = amount of acid in the final solution
40/100 * (60 + x) = 42
Multiplying both sides by 100, we get:
40 * (60 + x) = 4200
Expanding the equation:
2400 + 40x = 4200
Subtracting 2400 from both sides:
40x = 4200 - 2400
40x = 1800
Dividing both sides by 40:
x = 1800/40
x = 45
Therefore, the chemist should add 45 grams of water to make a solution that is 40% acid.
To solve this problem, we need to determine the quantity of water needed to dilute the 70% acid solution to a 40% acid solution.
First, let's calculate the amount of acid in the initial solution.
The chemist has 60 g of a 70% acid solution. So, the amount of acid in the initial solution is 0.7 * 60 = 42 grams.
Now, let's assume that x grams of water should be added to dilute the solution.
After adding x grams of water, the total mass of the solution will become 60 + x grams.
In the final solution, as it is desired for it to be a 40% acid solution, the amount of acid will be equal to 40% of the total mass of the solution.
So, the amount of acid in the final solution will be (40/100) * (60 + x) grams.
Since the amount of acid should remain constant before and after dilution, we can set up the equation:
42 grams = (40/100) * (60 + x) grams
Simplifying this equation, we have:
42 = 0.4 * (60 + x)
Solving for x, the amount of water to be added, we get:
42 = 24 + 0.4x
18 = 0.4x
x = 18 / 0.4
x = 45
Therefore, the chemist should add 45 grams of water to the 70% acid solution to obtain a 40% acid solution.