100mL solution of NaCl in water used 2.9g of salt. The final concentration is 0.50 M. express the concentration as a mass/volume %. How do these values compare to the molarity? Why? Show all calculations

A solution's mass by volume percent concentration, `"% m/v"`, is determined by looking at the number of grams of solute present in `"100 mL"` of solution.

In your case, `"2.9 g"` of sodium chloride, your solute, present in `"100 mL"` of solution would give you a `"2.9% m/v"` solution.

`100 color(red)(cancel(color(black)("mL solution"))) * overbrace("2.9 g NaCl"/(100color(red)(cancel(color(black)("mL solution")))))^(color(purple)("given by the problem")) = "2.9 g NaCl"`

Therefore, you have

`color(darkgreen)(ul(color(black)("% m/v " = " 2.9% NaCl")))`

As you know, the number of moles of solute is calculated by dividing its mass by its molar mass. In this case, sodium chloride has a molar mass of `"58.44 g mol"^(-1)`, which means that you have

`2.9 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) ~~ "0.050 moles NaCl"`

Now, molarity, `c`, is calculated by taking into account the number of moles of solute present in `"1 L"` of solution, so

`1color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.050 moles NaCl"/(100color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.50 M")))`

As you can see, the mass by volume percent concentration has a higher value than the solution's molarity.

Now, in order to get from molarity to mass by volume percent concentration, you must

• `color(blue)(ul(color(black)("divide the molarity by 10")))`
  This ensures that you're working with the number of moles of solute present in `"100 mL"` of solution
• `color(blue)(ul(color(black)("multiply the result by the molar mass of the solute")))`
  This ensures that you;'re working with the number of grams of solute present in `"100 mL"` of solution

So, let's try this out using a `"1.5-M"` solution of sodium chloride. You will have

`"1.5"/10 = 0.15 ->` the number of moles of solute present in `"100 mL"` of solution

`0.15 * 58.44 ~~ 8.8 ->` the number of grams of solute present in `"100 mL"` of solution

Therefore, you can say that

`color(darkgreen)(ul(color(black)("% m/v = 8.8% NaCl")))" "` and `" "color(darkgreen)(ul(color(black)(c_"NaCl" = "1.5 M")))`