Question

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?

Answer 1

`"0.1 moles NaCl"`

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 `"58.44 g"` of sodium chloride in `"1.0 L"` of water and make a `"1.0 M"` sodium chloride solution.

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. `"1.0 M"`.

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 `color(blue)(10)` times fewer moles of solute in order to have the same molarity.

Sodium chloride was said to have a molar mass of `"58.44 g mol"^(-1)`, which tells you that `1` mole of sodium chloride has a mass of `"58.44 g"`.

Since the initial solution was made by dissolving the equivalent of

`"58.44 g " = " 1 mole NaCl"`

in `"10 L"` of water, it follows that the `"0.1 L"` sample must contain

`"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 `"0.1 L"` sample contains

`"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 `n_'solute"` and plug in your values to find

`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.