Answer:
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.
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 = [tex]\rm \dfrac{weight}{molecular\;weight}\;\times\;\dfrac{1000}{volume\;(ml)}[/tex]
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 = [tex]\rm \dfrac{weight}{166.002}\;\times\;\dfrac{1000}{500}[/tex]
weight = [tex]\rm \dfrac{0.125\;\times\;166.002}{2}[/tex]
weight of KI = 10.37 grams.
For the preparation of 0.125 M of KI solution, 10.37 grams of KI is needed.
For more information about molarity, refer to the link:
https://brainly.com/question/16727614
