Question

how much water should be added to 110 oz. of 60% acid solution to dilute it to a 45% acid solution? round answer to the nearest hundredth if necessary

Answer 1

We are given the following information:

110 oz of 60% acid solution

We need to find the amount of water needed to dilute it to 45%.

To answer this, let's set up the equation first.

We know that 60% of the 110 oz. is acid. This amount of acid must then be the same amount of acid that makes up 45% of the new solution. So our equation would be:

$60\%\times110=45\%\times N$

Solving for N, we get:

$\begin{gathered} 0.60(110)=0.45N \\ 66=0.45N \\ \frac{66}{0.45}=N \\ \\ N=146.666...\approx146.67 \end{gathered}$

Now we know that the total amount must be 146.67 oz. We already have 110 oz. So we only need 146.67 - 110 = 36.67 oz of water to dilute the solution.