Question

calculate the volume in liters of a 316. g/l zinc nitrate solution that contains 414. g of zinc nitrate (zn(no₃)₂).
be sure your answer has the correct number of significant digits.

l

Explanation:

Step1: Use the formula for concentration

The formula for concentration \(C=\frac{m}{V}\), where \(C\) is concentration (\(g/L\)), \(m\) is mass (\(g\)), and \(V\) is volume (\(L\)). We can rearrange it to solve for \(V\): \(V = \frac{m}{C}\).

Step2: Substitute the given values

We are given \(m = 414\space g\) and \(C=316\space g/L\). Substituting these values into the formula: \(V=\frac{414\space g}{316\space g/L}\).

Step3: Calculate the volume

\(V=\frac{414}{316}\space L\approx1.31\space L\) (rounded to three significant digits as the least number of significant digits in the given values \(414\) (three) and \(316\) (three)).

Answer:

\(1.31\space L\)