# Question #8f49b

##### 1 Answer

#### Explanation:

The trick here is to focus on how the concentration of the target solution relates to the concentration of the initial solution.

You don't have to know the *exact values* of the two concentrations, all you need to know is that

#"N"/2 xx 1/color(red)(5) = "N"/10#

This means that the concentration of the target solution **decreased** by a factor of

Now, the key to any *dilution* calculation is the fact that when you're diluting a solution, the *number of moles* of solute **remains unchanged**.

In other words, you can **decrease** the concentration of a solution by **increasing its volume**, while at the same time keeping the number of moles of solute **constant**.

This means that in order for the concentration of a solution to **decrease** by a factor of **increase** by a factor of

Therefore, the volume of the target solution must be

#V_"target" = color(red)(5) xx "150 mL" = "750 mL"#

This means that you must add

#V_"water" = "750 mL" - "150 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("600 mL")color(white)(a/a)|)))#

of water to your initial solution to get its concentration to drop from

Mathematically, this can be shown by using the equation for *dilution calculations*

#color(blue)(overbrace(c_1 xx V_1)^(color(darkgreen)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(darkgreen)("moles of solute in diluted solution"))#

Here

Rearrange to solve for

#c_1 * V_1 = c_2 * V_2 implies V_2 = c_1/c_2 * V_1#

Plug in your values to find

#V_2 = color(red)(cancel(color(black)("N")))/2 * 10/color(red)(cancel(color(black)("N"))) * "150 mL" = "750 mL"#

Once again ,the volume of water that must be added is

#V_"water" = "750 mL" - "150 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("600 mL")color(white)(a/a)|)))#