# Question 27485

Aug 23, 2017

Here's what I got.

#### Explanation:

For starters, the density given to you is too high for a 50% $\text{m/m}$ potassium hydroxide solution. This solution should actually have a density of approximately ${\text{1.52 g cm}}^{- 3}$

https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-potassium-hydroxide/

I'll do the calculations with the actual value of the density, but feel free to redo the calculations with the value given to you if needed.

Now, assuming that your goal here is to figure out the molarity of the solution, start by picking a sample of this potassium hydroxide solution.

To make the calculations easier, let's pick a sample that has a volume of

${\text{1 dm}}^{3} = {10}^{3}$ ${\text{cm}}^{3}$

In order to find the molarity of the solution, you must find the number of moles of solute present in this sample.

Use the density of the solution to determine its mass

10^3 color(red)(cancel(color(black)("cm"^3color(white)(.)"solution"))) * "1.52 g"/(1color(red)(cancel(color(black)("cm"^3color(white)(.)"solution")))) = "1520 g"

The solution is said to be 50% "m/m" potassium hydroxide, which means that you will get $\text{50 g}$ of potassium hydroxide, the solute, for every $\text{100 g}$ of solution.

You can thus say that your sample will contain

1520 color(red)(cancel(color(black)("g solution"))) * "50 g KOH"/(100color(red)(cancel(color(black)("g solution")))) = "760 g KOH"

To convert the mass of potassium hydroxide to moles, use the molar mass of the compound

760 color(red)(cancel(color(black)("g"))) * "1 mole KOH"/(56.11color(red)(cancel(color(black)("g")))) = "13.54 moles KOH"#

Since this represents the number of moles of solute present in ${10}^{3} \textcolor{w h i t e}{.} {\text{cm"^3 = "1 dm}}^{3}$ of solution, you can say that the molarity of the solution will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 14 mol dm}}^{- 3}}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the percent concentration of the solution.