Question

How do you find the molarity of a salt solution made by dissolving 280 mg of NaCl in 2.00 mL of solution?

Answer 1

`"2.4 M"`

Explanation:

In order to be able to determine the molarity of a solution, you basically need to know two things

• how many moles of solute you have in your sample
• the volume of this sample - expressed in liters

Molarity is defined as

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

Your solute is sodium chloride, `"NaCl"`. You know that you're dissolving a `"280-mg"` sample of sodium chloride in enough water to get a `"2.00-mL"` volume of solution.

Notice that the problem provides you with the volume of the solution, but that it is expressed in milliliters. This means that you will have to convert it to liters by using the conversion factor

`"1 L" = 10^3"mL"`

The volume of the solution will thus be

`2.00 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 2.00 * 10^(-3)"L"`

Now, in order to get the number of moles of solute, you need to use sodium chloride's molar mass, which tells you the mass of one mole of this compound.

Sodium chloride has a molar mass of `"58.44 g/mol"`, which tells you that one mole of sodium chloride has a mass of `"58.44 g"`.

Notice that the sample you have is expressed in milligrams. Once again, a unit conversion is in order.

`280 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3color(red)(cancel(color(black)("mg")))) = 280 * 10^(-3)"g"`

This means that your sample will contain

`280 * 10^(-3)color(red)(cancel(color(black)("g NaCl"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g NaCl")))) = 4.79 * 10^(-3)"moles NaCl"`

Therefore, the molarity of this solution will be

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

`c = (4.79 * color(purple)(cancel(color(black)(10^3)))"moles")/(2.00 * color(purple)(cancel(color(black)(10^3)))"L") = "2.395 M"`

Rounded to two sig figs, the number of sig figs you have for the mass of sodium chloride, the answer will be

`c = color(green)("2.4 M")`

SIDE NOTE Because sodium chloride is soluble in aqueous solution, which means that it exists as sodium cations, `"Na"^(+)`, and chloride anions, `"Cl"^(-)`, you wouldn't actually say that you have a `"2.4-M"` sodium chloride solution.

Instead, you would say that you have a solution that is `"2.4 M"` in `"Na"^(+)` and `"2.4 M"` in `"Cl"^(-)`.