# If molarity moles/liter, then how many grams of NaCl are needed to make 3.550 liter of a 1.30 M solution?

Nov 22, 2016

$\text{270. g}$

#### Explanation:

Molarity is indeed defined as moles of solute per liter of solution, but it can also be expressed as grams of solute per liter of solution.

As you know, a $\text{1 M}$ solution contains exactly $1$ mole of solute in $\text{1 L}$ of solution.

In your case, a $\text{1.30 M}$ sodium chloride solution will contain $1.30$ moles of sodium chloride, the solute, for every $\text{1 L}$ of solution.

You can use the molar mass of sodium chloride to convert this to grams of sodium chloride for every $\text{1 L}$ of solution.

1.30 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "75.97 g"

You can now say that a $\text{1.30 M}$ sodium chloride solution must contain $\text{75.97 g}$ of sodium chloride for every $\text{1 L}$ of solution. Now all you have to do is use this concentration to figure out how many grams would be needed to make your sample

$3.550 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L solution"))) * "75.97 g NaCl"/(1color(red)(cancel(color(black)("L solution")))) = color(darkgreen)(ul(color(black)("270. g NaCl}}}}$

The answer is rounded to three sig figs.