Question

A 1.200 liter solution contains 35.0 grams of potassium iodide (KI). What is the molarity of this

solution?

Answer 1

M = 0.177

Step-by-step explanation:

First, we have to find the molar mass of potassium iodine (KI).

K = 39.098

I = 126.904

Now, add these values together to get the molar mass of KI

39.098 + 126.904 = 166.0028

Now, it's time to do a grams to moles conversion.

35.0 g KI *
`(1 mol KI)/(166.0028 g KI)` =
`(35.0)/(166.0028)` = 0.212 mol KI

Now, we can find the molarity of this solution.

Molarity (M) =
`(mol)/(L)`

M =
`(0.212 mol KI)/(1.200 L)` = 0.177 M

The molarity (M) of this solution is 0.177.