Question #5ab95

2 Answers
Jul 16, 2015

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.

#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.

Jul 16, 2015

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:
#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.#