# What is the molarity of a salt solution made by dissolving 280.0 mg of NaCl in 2.00 mL of water?

## Assume the final volume is the same as the volume of the water.

Feb 29, 2016

$\text{2.40 M}$

#### Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters, $\text{mL}$. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

$\text{1 L" = 10^3"mL}$

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol, ${\text{g mol}}^{- 1}$, which means that you're going to have to convert the mass of the sample from milligrams to grams

$\text{1 g" = 10^3"mg}$

Sodium chloride, $\text{NaCl}$, has a molar mass of ${\text{58.44 g mol}}^{- 1}$, which means that your sample will contain

overbrace(280.0 color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))))^(color(purple)("unit conversion")) * overbrace("1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))))^(color(brown)("molar mass")) = "0.004791 moles NaCl"

This means that the molarity of the solution will be

$\textcolor{b l u e}{c = {n}_{\text{solute"/V_"solution}}}$

c = "0.004791 moles"/(2.00 * 10^(-3)"L") = color(green)("2.40 M")

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.