Question

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?

Answer 1

`"3.00 mol L"^(-1)`

Explanation:

Molarity is a measure of how many moles of solute you get in exactly `"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 `"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 `"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

`color(darkgreen)(ul(color(black)("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.