Question

A chemist adds 370.0mL of a 1.41/molL potassium iodide KI solution to a reaction flask. Calculate the millimoles of potassium iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer 1

As we know moalrity of a solution is given by,

Molarity = No. of moles of substance / Volume of solution

Also, we are given

Volume of KI (Potassium Iodide) solution = 370 ml = 0.370 L               (3 significant figures)

Molarity of KI solution = 1.41 mol / L                                                 (3 significant figures)      

On substituting given values we get,

No. of moles of KI (Potassium Iodide) in solution = Volume * Molarity of Solution

No. of moles of KI (Potassium Iodide) in solution = (0.370 L) * (1.41 mol/L) = 0.5217 mol

No. of moles of KI = 0.5217 moles = 521.7 millimoles = 522 millimoles (rounded off to 3 significant figures)

NOTE : ( 1 mole = 1000 millimole)