# You want to produce 579 mL of a 0.440 M solution of NaCl. You have a 1.40 M solution. How many mL of it should you use and dilute with water?

##### 1 Answer

#### Answer:

#### Explanation:

Let's start by **assuming** that you're not familiar with the formula for dilution calculations, so that you can't just plug in your values and do a quick calculation.

So, the idea with dilution calculations is that the number of moles of solute **remains constant** when diluting a given sample.

In essence, to dilute a solution you need to **decrease** the concentration of the solute by keeping the number of moles of solute *constant* and **Increasing** the volume.

As you know, molarity is defined as moles of solute, which in your case will be sodium chloride, **liters** of solution.

#color(blue)(c = n/V)#

Use the molarity and volume of the target solution to determine how many moles of solute it must contain

#color(blue)(c = n/V implies n = c * V)#

#n_(NaCl) = "0.440 M" * 579 * 10^(-3)"L" = "0.25476 moles NaCl"#

This is exactly how many moles of solute must be present in the **sample of stock solution** used to prepare the target solution.

This means that you can use the definition of molarity to determine what *volume* of the stock solution would contain this many moles of sodium chloride

#color(blue)(c = n/V implies V = n/c)#

#V = (0.25476 color(red)(cancel(color(black)("moles"))))/(1.40 color(red)(cancel(color(black)("moles")))/"L") = "0.182 L"#

Expressed in *milliliters*, the answer will be

#V_"stock" = color(green)("182 mL")#

**Alternatively**, you can use the formula for dilution calculations, which expresses the exact same concept

#overbrace(c_1V_1)^(color(purple)(stackrel("moles of solute")("in stock solution"))) = overbrace(c_2V_2)^(color(red)(stackrel("moles of solute")("in target solution")))" "# , where

In this case, you would get

#V_1 = c_2/c_1 * V_2#

#V_1 = (0.440 color(red)(cancel(color(black)("M"))))/(1.40color(red)(cancel(color(black)("M")))) * "579 mL" = color(green)("182 mL")#