Question

Explain how to prepare 100ml of a 1.0 m ki solution

Answer 1

You will need to calculate the mass of KI you will need to dissolve in water in order to prepare the solution. The mass could be calculated by converting 100 ml to L then multiplying that to the target molarity of the solution which is 1.0 m. Then multiply it with the molecular weight of KI which is 166 g. Thus, you will need to dissolve 16.6 grams of KI to 100 ml water. 

Answer 2

Answer:

Weight 16.6 g of KI, dissolve it in water, transfer to a volumetric ball of 100 mL and complete the volume with water.

Explanation:

You must be careful with concentrations. First we have molarity, represented by a capital M, and then we have molality, represented by a lowercase m; they are defined next:

Molarity = M = moles of solute/ Liters of solution

Molality = m = moles of solute / Kg of solvent

if we were talking about molality, the problem wouldn’t be so easily resolve since they don´t give us the mass of solvent to be use but the volume of solution need and you will need the solution’s density to have an approximation of the mass solvent.

For this reason, I’m assuming we are talking about molarity. To solve the problem, we need to find the amount of KI we need to prepare the solution, for this we use the equation of molarity:

M = moles KI / Liters of solution

With this we can find the moles of KI needed:

Moles KI = M x liters of solution

Replacing the values:

Moles KI = 1.0 M x 0.1 L = 0.1 moles of KI

Note that 100 mL = 0.1 L. Now we need the grams of KI to be weighted. Using its molecular mass, which is 166 g/mol, we have:

$0.1 mole KI \frac{166 g KI}{1molKI} = 16.6 g KI$

Then, we need to weight 16.6 g of KI, dissolve it in water (about 50 mL), transfer to a volumetric ball of 100 mL and complete the volume with water.