Question

2. How many grams of potassium nitrate are required to prepare 0.250 L of a 0.350 M solution?

Answer 1

Answer

8.847 grams

Explanation

Given:

Volume of the solution = 0.250 L

Molarity of the solution = 0.350 M

What to find:

The grams of potassium nitrate required to prepare the solution.

Step-by-step solution:

The solution involves two steps. Step one is to determine the moles of potassium nitrate present in 0.350 M solution and the final step is to convert the moles to grams of potassium nitrate.

Step 1: Determine the moles of potassium nitrate in 0.350 M solution using the molarity formula.

$\begin{gathered} Molarity=\frac{Mole}{Volume\text{ }in\text{ }L} \\ \\ Mole=Molarity\times Volume\text{ }in\text{ }L \end{gathered}$

So, Mole = 0.350 M x 0.250 L = 0.0875 mol

Step 2: Convert 0.0875 mol KNO₃ to grams using the mole formula.

Molar mass of KNO₃ = 101.1032 g/mol

$\begin{gathered} Mole=\frac{Mass}{Molar\text{ }mass} \\ \\ \Rightarrow Mass=Mole\times Molar\text{ }mass \end{gathered}$

Mass of KNO₃ = 0.0875 mol x 101.1032 g/mol = 8.847 grams.

Hence, 8.847 grams of potassium nitrate are required to prepare 0.250 L of a 0.350 M solution.