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

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

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

##### 1 Answer

#### Answer:

#### 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*, **must** convert this volume to *liters* by using the conversion factor

#"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*, *milligrams* to *grams*

#"1 g" = 10^3"mg"#

Sodium chloride,

#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

#color(blue)(c = n_"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.