Question

a chemist adds 1.20 l of a 0.0551m potassium iodide (ki) solution to a reaction flask. calculate the moles of potassium iodide the chemist has added to the flask. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall molarity formula

Molarity ($M$) = $\frac{\text{moles of solute}(n)}{\text{volume of solution in liters}(V)}$, so $n = M\times V$.

Step2: Substitute given values

Given $M = 0.0551\ M$ and $V=1.20\ L$. Then $n=0.0551\ mol/L\times1.20\ L$.

Step3: Calculate the result

$n = 0.0551\times1.20=0.06612\ mol$. Rounding to three - significant digits (since 0.0551 and 1.20 have three significant digits), we get $n = 0.0661\ mol$.

Answer:

$0.0661\ mol$