2. Calculate the number of grams of potassium iodide (KI), needed to prepare 500.0 mL of a 0.125 M KI. Show your work.

For the preparation of 0.125 M of KI solution, 10.37 grams of KI is needed.

Molarity can be defined as the moles of solute in a liter of solution.

Molarity =

Given, Molarity = 0.125 M

volume = 500 ml

We know the molecular weight of KI = 166.002 g/mol

Putting the values in the equation:

0.125 M =

weight =

weight of KI = 10.37 grams.

For the preparation of 0.125 M of KI solution, 10.37 grams of KI is needed.

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10.38 g

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 500 mL

Molarity of KI solution = 0.125 M

Mass of KI needed =?

Next, we shall convert 500 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Thus, 500 mL is equivalent to 0.5 L.

Next, we shall determine the number of mole KI in the solution. This can be obtained as illustrated below:

Volume of solution = 0.5 L

Molarity of KI solution = 0.125 M

Mole of KI =?

Molarity = mole /Volume

0.125 = mole of KI /0.5

Cross multiply

Mole of KI = 0.125 × 0.5

Mole of KI = 0.0625 mole

Finall, we shall determine the mass of KI needed to prepare the solution as follow:

Mole of KI = 0.0625 mole

Molar mass of KI = 39 + 127 = 166 g/mol

Mass of KI =?

Mole = mass /Molar mass

0.0625 = mass of KI /166

Cross multiply

Mass of KI = 0.0625 × 166

Mass of KI = 10.38 g

Therefore, 10.38 g is needed to prepare the solution.

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