Question

calculate the volume in liters of a 0.23m potassium permanganate solution that contains 100. g of potassium permanganate (kmno₄). be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate molar mass of KMnO₄

The molar mass of K (potassium) is approximately 39.1 g/mol, Mn (manganese) is about 54.9 g/mol, and O (oxygen) is 16.0 g/mol. For KMnO₄, molar mass $M=(39.1 + 54.9+4\times16.0)$ g/mol = 158.0 g/mol.

Step2: Calculate moles of KMnO₄

Use the formula $n=\frac{m}{M}$, where $m = 100. g$ and $M = 158.0$ g/mol. So $n=\frac{100.}{158.0}$ mol $\approx0.633$ mol.

Step3: Calculate volume of the solution

Use the formula for molarity $M=\frac{n}{V}$, where $M = 0.23$ M and $n=0.633$ mol. Rearranging for $V$, we get $V=\frac{n}{M}=\frac{0.633}{0.23}$ L $\approx2.75$ L.

Answer:

2.8 L