How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?

1 answer:

8 0

Suppose you add x liters of pure water to the 10 L of 25% acid solution. The new solution's volume is x + 10 L. Each L of pure water contributes no acid, while the starting solution contains 2.5 L of acid. So in the new solution, you end up with a concentration of (2.5 L)/(x + 10 L), and you want this concentration to be 10%. So we have

$\dfrac{2.5}{x+10}=0.1\implies25=x+10\implies x=15$

and so you would need to add 15 L of pure water to get the desired concentration of acid.

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How do you solve this whole thing please?

NemiM [27]

Answer:

Step-by-step explanation:

Let's look at the first two, and hopefully you'll be able to figure the rest out:

1. Answer below

$d = rt$

The problem asks us to solve for $r$ , so that means we need to get $r$ on one side of the equation by itself. To do so, we will need to divide both sides of the equation by $t$ :

$\frac{d}{t} = \frac{rt}{t}$

$\frac{d}{t} = r$

2. Answer below

$B = T - Lc$

The problem asks us to solve for $T$ , so that means we need to get $T$ on one side of the equation by itself. To do so, we will need to add $Lc$ to both sides of equation: