Question

Question #5ab95

Answer 1

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 = n/V => n = C * V`

`n_(KIO_4) = 0.03 "moles"/cancel("L") * 1cancel("L") = "0.03 moles"` `KIO_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"` `KIO_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.

Answer 2

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:
`KIO_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.`