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

Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.

Explanation:

Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L

Molarity of KI solution = 1.41 mol/L

Now, moles of KI (potassium iodide) is calculated as follows.

$Moles = Volume \times Molarity \\= 0.37 L \times 1.41 M\\= 0.5217 mol$

Convert moles into millimoles as follows.

1 mol = 1000 millimoles

0.5217 mol = $0.5217 mol \times \frac{1000 millimoles}{1 mol} = 521.7 millimoles$

This can be rounded off to the value 522 millimoles.

Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.