# Question #388f4

##### 1 Answer

#### Answer:

That depends on what exactly you're dealing with.

#### Explanation:

If I understand your question correctly, you're dealing with a solution that is has a percent concentration of **3.48%**.

Now, let's assume that this percent concentration is **weight by volume**, **w/v**.

In simple terms, a **3.48% w/v** solution of something contains **3.48 g** of that something in *every* 100 mL of solution. If that is the case, 10 mL of this solution would contain **10 times fewer** grams

If you were to take a **1000-mL** sample of this olution, than it would contain

The volume is *100 times bigger*, the solution will contain *100 times more* solute.

Now, let's assume that you want to perform a *dilution* of the 10-mL solution and increase its volume to **1000-mL**.

In this case, you would add enough water to the first solution to get the total volume to 1000 mL - you'd add 990 mL of water, to be precise.

The solution *would still contain* **0.348 g** of something, since you only added water, but its weight by volume percent concentration would be

This **new solution** wil contain **0.0348 g** of something for *every* **100 g** of solution.