Question

a chemist prepares a solution of iron(iii) bromide (febr₃) by measuring out 1.3×10² μmol of iron(iii) bromide into a 350. ml volumetric flask and filling the flask to the mark with water. calculate the concentration in μmol/l of the chemists iron(iii) bromide solution. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert volume to liters

The volume of the solution is $V = 350\ mL$. Since $1\ L=1000\ mL$, then $V = 350\ mL\times\frac{1\ L}{1000\ mL}=0.350\ L$.

Step2: Calculate the concentration

The formula for concentration $C$ is $C=\frac{n}{V}$, where $n = 1.3\times10^{2}\ \mu mol$ is the amount - of - substance and $V$ is the volume of the solution. Substitute $n = 1.3\times10^{2}\ \mu mol$ and $V = 0.350\ L$ into the formula: $C=\frac{1.3\times 10^{2}\ \mu mol}{0.350\ L}\approx371.43\ \mu mol/L$. Rounding to two significant digits (because 1.3 has two significant digits), $C = 3.7\times10^{2}\ \mu mol/L$.

Answer:

$3.7\times 10^{2}\ \frac{\mu mol}{L}$