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")))#