# What is the molarity of a "KOH" solution with relative density 0.0028 and %"w/w" = 40%? The molar mass of "KOH" is "56.1056 g/mol".

Aug 23, 2017

$c = \text{0.02 M}$

The molarity of a solution is given by

$c = \text{mols solute"/"L solution}$,

so we'll need to get the solution's volume and the mols of solute.

The relative density, or specific gravity, is the density of the solution relative to that of water (it would have to be based on the solution; if it was based on the solid, it would be much higher), i.e.

${\rho}_{\text{rel" = 0.0028 = rho_"soln}} / {\rho}_{w}$

Since the density of water is close to $\text{1 g/mL}$, the density of the solution is about $\text{0.0028 g/mL}$, or $\text{2.8 g/L}$. Knowing that the percent by mass is

"%w/w" = "mass of solute"/"mass of solution" xx 100%,

we then know that for every $\text{100 g}$ of solution, there is $\text{40 g}$ of solute. Since density is an extensive property, it doesn't matter what mass of solution we choose, so let's choose $\text{100 g}$.

Using the density of the solution, its volume is:

$\underline{{V}_{\text{soln") = 100 cancel"g solution" xx "1 L soln"/(2.8 cancel"g soln}}}$

$=$ $\underline{\text{35.7 L soln}}$

Using the molar mass of $\text{KOH}$, we have this many mols of it:

$\underline{{n}_{\text{KOH") = 40 cancel"g KOH" xx "1 mol KOH"/(56.1056 cancel"g KOH}}}$

$=$ $\underline{\text{0.713 mols KOH}}$

Therefore, the molarity is:

$\textcolor{b l u e}{c} = \text{0.713 mols KOH"/"35.7 L soln}$

$=$ $\textcolor{b l u e}{\text{0.02 M}}$