# Given that the molar mass of NaCl 58.44 g/mol, what is the molarity of a solution that contains 87.75 g of NaCl in 500. mL of solution?

Feb 24, 2017

${\text{3.00 mol L}}^{- 1}$

#### Explanation:

Molarity is a measure of how many moles of solute you get in exactly $\text{1 L}$ of a given solution.

Your goal here is to use the molar mass of sodium chloride to find the number of moles present in the sample.

87.75 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.5015 moles NaCl"

Now, you know that this solution contains $1.5015$ moles of sodium chloride, the solute, in $\text{500. mL}$ of solution. Since

500.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.500 L"

you can use this known composition of the solution as a conversion factor to help you find the number of moles of solute present in $\text{1 L}$ of solution

1 color(red)(cancel(color(black)("L solution"))) * "1.5015 moles NaCl"/(0.500color(red)(cancel(color(black)("L solution")))) = "3.003 moles NaCl"

You can now say that the molarity of the solution is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 3.00 mol L}}^{- 1}}}}$

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