# Question #f3cad

##### 1 Answer

#### Explanation:

I will assume that the glucose solution **is being added** to

The idea here is that the concentration of the solution will **decrease** because the mass of solute *remains constant* while the volume of solvent **increases**.

Your initial solution had a concentration of **for every** **of solution**.

Now, you don't have to calculate the amount of solute present in the initial solution, all you have to do here is figure out what the **total volume** of the solution will be after the dilution.

#V_"total" = "25 mL" + "375 mL" = "400 mL"#

You can say that the initial solution is being diluted by a factor of

#"DF" = (400 color(red)(cancel(color(black)("mL"))))/(25color(red)(cancel(color(black)("mL")))) = color(blue)(16) -># this is thedilution factor

This means that the diluted solution contains **times more solvent** and the **same amount of solute** as the initial solution contained.

Therefore, you can say that the concentration of the diluted solution is **times lower** than that of the initial solution

#"% diluted" = "15%"/color(blue)(16) = color(darkgreen)(ul(color(black)(0.94%)))#

The answer is rounded to two **sig figs**.