# Question #5ab95

##### 2 Answers

#### Answer:

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.

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.

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.*

#### Answer:

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:

One mol of potassium (

One mol of iodine (

Four mol of oxygen (

Summing up we get

Provided a solution 0.03 M contains 0.03 moles per liter, we calulate