# Question db8f3

Jan 13, 2015

I assume you need to determine the $\text{w/v}$ percent concentration (or m/v) and the $\text{w/w}$ percent concentration (or m/m); since I'm not really sure what m% means, I'll solve for molality as well, just to be on the safe side.

So, in order to get the $\text{w/v}$ percent concentration we need the mass of the solute and the total volume of the solution. We know that the solution has a molarity of ${\text{0.83 mol/dm}}^{3}$; this will allow you to determine how many moles of $N {a}_{2} S {O}_{4}$ are in 1.0 L of solution

$C = \frac{n}{V} \implies {n}_{N {a}_{2} S {O}_{4}} = C \cdot V$

${n}_{N {a}_{2} S {O}_{4}} = 0.83$ ${\text{moles"/("dm"^3) * "1 dm}}^{3} = 0.83$ $\text{moles}$

You can now determine the mass of $N {a}_{2} S {O}_{4}$ by multiplying the number of moles by $\text{142.0 g/mol}$, the compound's molar mass

${m}_{N {a}_{2} S {O}_{4}} = \text{0.83 moles" * "142.0}$ $\text{g"/"mol} = 117.9$ $\text{g}$

So, your $\text{w/v}$ percent concentration wil be

"w/v%" = ("117.9 g")/("1.0 L") * 100% = 11.8%

Now, if you are looking for the $\text{w/w}$ percent concentration, you have to determine what the mass of the solution is; this is where density comes into play. For the same 1.0 L sample,

${\text{1.1" "g"/("dm"^3) * "1000 dm}}^{3} = 1100$ $\text{g solution}$

This means that

"m/m%" = ("117.9 g")/("1100 g") * 100% = 10.7%

The solution's molality requires the mass of the solvent in kg, which in this case I assume is water. SInce the total mass of the solution is 1100 g, and the solute weighs 117.9 g, the mass of water is

m_("water") = "1100 g" - "117.9 g" = "982.1 g", therefore

"m" = ("0.83 moles Na"_2"SO"_4)/("982.1" * "10"^(-3) "kg") = 0.85# $\text{molal}$