Question #5ab95
2 Answers
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.
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