Question

how much potassium iodide (KI) in grams should be added to water to produce 3.5L of solution with a. concentration of 0.75 M ?

Answer 1

The mass of KI is 435.76 grams

Step-by-step explanation

Given that;

The volume of the solution is 3.5L

The concentration of the solution is 0.75M

Follow the steps below to find the number of moles of KI

Step 1; Write the molarity formula

`\text{ Molarity = }\frac{\text{ moles of solute}}{\text{ volume of solution}}`

`\begin{gathered} \text{ 0.75 = }\frac{\text{ moles of solute}}{\text{ 3.5}} \\ \text{ cross multiply} \\ \text{ moles of solute = 0.75 }*\text{ 3.5} \\ \text{ moles of solute = 2.625 mol} \end{gathered}`

Step 2; Find the mass of the solute using the formula below

`\begin{gathered} \text{ moles = }\frac{\text{ mass}}{\text{ molar mass}} \\ \text{ cross multiply} \\ \text{ mass = moles }*\text{ molar mass} \end{gathered}`

Recall, that the molar mass of KI is 166.0028 g/mol

`\begin{gathered} \text{ mass = 2.625 }*\text{ 166.0028} \\ \text{ mass = 435.76 grams} \end{gathered}`

Therefore, the mass of KI is 435.76 grams