# Question 5ab95

Jul 16, 2015

You dissolve 0.03 moles of potassium periodate in enough water to make one liter of solution.

#### Explanation:

A solution's molarity is defined as the number of moles of solute, in your case potassium periodate, per liter of solution.

So, a 0.03-M solution would contain 0.03 moles of solute for every 1 liter of solution.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{K I {O}_{4}} = 0.03 \text{moles"/cancel("L") * 1cancel("L") = "0.03 moles}$ $K I {O}_{4}$

To determine how many grams of potassium periodate would contain this many moles, use the compound's molar mass, which tells you what the mass of 1 mole of a substance is.

0.03cancel("moles") * "230.0 g"/(1cancel("mole")) = "6.9 g" $K I {O}_{4}$

If you take into account the number of sig figs given for the molarity of the solution, then you can prepare your target solution by dissolving 7 grams of potassium periodate in enough water to make one liter.

SIDE NOTE Something I forgot to mention - at room temperature, potassium periodate has a molar solubility of about 0.02 mol/L.

This means that, in order to be able to dissolve 0.03 moles in one liter, you're going to have to increase the temperature of the water a bit.

At room temperature, the maxiumum molarity of a potassium periodate solution is 0.02 M.

Jul 16, 2015

Just dissolve 6.9 grams of potassium periodate in deionized water up to make exactly one liter to obtain one liter of 0.03 M solution. If you need more, or less volume, simply calculate from this.

#### Explanation:

For example, if you need one tenth of one liter, 100 mL, weigh 0.69 grams of white powdered potassium periodate (i.e. one tenth of 6.9 grams) and put it in a flask of 100 mL (one tenth of a Liter).

Thee are the calculations:
$K I {O}_{4}$ is named potassium periodate. One mole of this compound contains:
One mol of potassium ($K$) = $39.1$ grams
One mol of iodine ($I$) = $126.9$ grams
Four mol of oxygen (${O}_{4}$) = 4·16.0 = 64.0 grams

Summing up we get $39.1 g + 126.9 g + 64.0 g = 230.0 g$ as the weight of a mol.

Provided a solution 0.03 M contains 0.03 moles per liter, we calulate 230 g/(mol) · 0.03 mol = 6.9 g.#