# 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

## mass/volume % formula is mass of solute (g) / mass of solvent (mL)

Dec 11, 2016

$\text{% m/v = 2.9% NaCl}$

#### Explanation:

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

In your case, $\text{2.9 g}$ of sodium chloride, your solute, present in $\text{100 mL}$ of solution would give you a $\text{2.9% m/v}$ 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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% m/v " = " 2.9% NaCl}}}}$

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 ${\text{58.44 g mol}}^{- 1}$, 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 $\text{1 L}$ of solution, so

$1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.050 moles NaCl"/(100color(red)(cancel(color(black)("mL solution")))) = color(darkgreen)(ul(color(black)("0.50 M}}}}$

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

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{divide the molarity by 10}}}}$
This ensures that you're working with the number of moles of solute present in $\text{100 mL}$ of solution
• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{multiply the result by the molar mass of the solute}}}}$
This ensures that you;'re working with the number of grams of solute present in $\text{100 mL}$ of solution

So, let's try this out using a $\text{1.5-M}$ solution of sodium chloride. You will have

$\frac{\text{1.5}}{10} = 0.15 \to$ the number of moles of solute present in $\text{100 mL}$ of solution

$0.15 \cdot 58.44 \approx 8.8 \to$ the number of grams of solute present in $\text{100 mL}$ 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")))