Question

How many grams of potassium iodide would you need to make 2 liters of a 0.075 M KI solution given that the molecular weight of KI=166 grams/mole?

Answer 1

24.9 grams of potassium iodide.

Step-by-step explanation:

We have the molarity of KI solution which is 0.075 M and the volume which is 2 liters (L). Based on these data, we can find the number of moles of KI, by using the molarity formula:

`Molarity\text{ \lparen M})=\frac{mole\text{s of solute}}{liters\text{ of solution}}=(mol)/(L).`

We can solve for moles of solute, like this:

`mole\text{s of solute}=Molarity\cdot liter\text{s of solution.}`

And then, we replace the values that we have in the new formula:

`\begin{gathered} moles\text{ of solute}=0.075M\cdot2L, \\ moles\text{ of solute}=0.15\text{ moles KI.} \end{gathered}`

Finally, we have to do the conversion from 0.15 moles of KI to grams using its molecular weight which you can see in the statement of the question (166 g/mol). The conversion will be:

`0.15\text{ moles KI}\cdot\frac{166\text{ g KI}}{1\text{ mol KI}}=24.9\text{ g KI.}`

We need 24.9 grams of potassium iodide to make 2 liters of a 0.075 M KI solution.