# The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt in 1.0 liter of solution. How many moles of NaCl would you have in 100 mL of this solution?

##### 1 Answer

#### Answer:

#### Explanation:

The thing to remember about **any solution** is that the particles of *solute* and the particles of *solvent* are **evenly mixed**.

This essentially means that if you start with a solution of known *molarity* and take out a sample of this solution, the molarity of the sample **will be the same** as the molarity of the initial solution.

In your case, you dissolve

If you take

#100 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.1 L"#

of this solution, this sample must have **the same concentration** as the initial solution, i.e.

Notice that the volume of the sample is

#(1color(red)(cancel(color(black)("L"))))/(0.1color(red)(cancel(color(black)("L")))) = color(blue)(10)#

**times** smaller than the volume of the initial solution, so it follows that it must contain **times** fewer moles of solute in order to have the *same molarity*.

Sodium chloride was said to have a **molar mass** of **mole** of sodium chloride has a mass of

Since the initial solution was made by dissolving the equivalent of

#"58.44 g " = " 1 mole NaCl"#

in

#"1 mole NaCl"/color(blue)(10) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))#

This is equivalent to saying that the

#"58.44 g NaCl"/color(blue)(10) = "5.844 g NaCl"#

**ALTERNATIVELY**

You can get the same result by using the formula for **molarity**, which is

#color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution"color(white)(a/a)|)))#

Rearrange this equation to solve for

#c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"#

#n_"NaCl" = "1.0 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.1 moles NaCl")color(white)(a/a)|)))#

Keep in mind that the volume must **always** be expressed in *liters* when working with molarity.