Question

calculate the volume in liters of a 33.5 g/dl zinc nitrate solution that contains 45.0 g of zinc nitrate (zn(no₃)₂). be sure your answer has the correct number of significant digits. l

Explanation:

Step1: Convert density units

Since \(1\ \text{dL} = 0.1\ \text{L}\), then the density \(33.5\ \text{g/dL}=335\ \text{g/L}\) (using unit conversion factor \(\frac{1\ \text{dL}}{0.1\ \text{L}}\)).

Step2: Use the formula \(V=\frac{m}{

ho}\)
We know that the mass \(m = 45.0\ \text{g}\) and the density \(
ho=335\ \text{g/L}\). Substitute into the formula \(V=\frac{m}{
ho}\), we get \(V=\frac{45.0\ \text{g}}{335\ \text{g/L}}\).

Step3: Calculate the volume

\(V=\frac{45.0}{335}\ \text{L}\approx0.134\ \text{L}\) (rounded to three significant digits as \(45.0\) has three significant digits and \(335\) has three significant digits).

Answer:

\(0.134\ \text{L}\)