Question

What is the molarity of a solution prepared by dissolving 37.94 g of potassium hydroxide, KOH, in some water and then diluting the solution to a volume of 500.00 mL? {Ans. ~ [KOH] = 1.352 M} Write the set up of the calculations that justify the answer.

Answer 1

1.352 M.

Step-by-step explanation:

What is given?

Mass of KOH = 37.94 g.

Volume of KOH solution = 500.00 mL.

Molar mass of K = 39.1 g/mol.

Molar mass of O = 16 g/mol.

Molar mass of H = 1 g/mol.

Step-by-step solution:

First, let's see the formula of molarity:

`Molarity\text{ \lparen M\rparen=}\frac{mole\text{s of solute}}{liter\text{s of solution}}=(mol)/(L).`

We have to find the moles of solute, our solute, in this case, is KOH. To convert 37.94 g of KOH to moles, we use the molar mass of KOH which is 56.10 g/mol using the given molar masses of the elements that can be found in the periodic table (39.1 g/mol + 16 g/mol + 1 g/mol = 56.1 g/mol). The conversion from grams to moles will look like this:

`37.94\text{ g KOH}\cdot\frac{1\text{ mol KOH}}{56.1\text{ g KOH}}=0.6763\text{ moles KOH.}`

And now, we have to find the volume in liters because we have the volume in milliliters. Remember that 1 L equals 1000 mL, so 500.0 mL in L is:

`500.0mL\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.5000\text{ L.}`

The final step is to replace the obtained values in the molarity formula:

`Molarity=\frac{0.6763\text{ moles}}{0.5000\text{ L}}=1.352\text{ M.}`

The molarity of solution would be 1.352 M.