Question

calculate the volume in milliliters of a 0.77 mol/l iron(ii) bromide solution that contains 225. mmol of iron(ii) bromide ($febr_2$). round your answer to 2 significant digits.

Explanation:

Step1: Convert mmol to mol

$225. \text{ mmol} = 225. \times 10^{-3} \text{ mol} = 0.225 \text{ mol}$

Step2: Use molarity formula to find volume

Molarity $M = \frac{n}{V}$, rearrange to $V = \frac{n}{M}$
$V = \frac{0.225 \text{ mol}}{0.77 \text{ mol/L}} \approx 0.2922 \text{ L}$

Step3: Convert L to mL

$0.2922 \text{ L} \times 1000 = 292.2 \text{ mL}$

Step4: Round to 2 significant digits

$292.2 \text{ mL} \approx 290 \text{ mL}$

Answer:

290 mL