# How do you calculate the mass of NaCl required to prepare 0.5 liters of a 2.5 molar solution of NaCl?

Dec 26, 2016

Here's how you can do that.

#### Explanation:

Your starting point here will be the definition of molarity.

As you know, molarity is defined as the number of moles of solute present in exactly $\text{1 L}$ of solution. This implies that a $\text{1-M}$ solution will contain $1$ mole of solute in $\text{1 L}$ of solution.

In your case, a $\text{2.5-M}$ solution will contain $2.5$ moles of sodium chloride, your solute, for every $\text{1 L}$ of solution. It follows that this sample must contain

0.5 color(red)(cancel(color(black)("L solution"))) * overbrace("2.5 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("required molarity")) = "1.25 moles NaCl"

Now, to convert this to grams of sodium chloride, you must use the mass of $1$ mole of this compound as a conversion factor. The mass of $1$ mole of sodium chloride is given by its molar mass

$1.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(darkgreen)(ul(color(black)("73 g}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the solution.