Question

What mass of \(\mathrm{KNO}_{3}\) is required to prepare \(150 . \mathrm{mL}\) of \(0.250 M \mathrm{KNO}_{3}\) solution?

Short answer

The mass of \(\mathrm{KNO}_{3}\) required to prepare 150 mL of 0.250 M \(\mathrm{KNO}_{3}\) solution is approximately 3.79 g.

Step-by-step solution

Find the molecular weight (molar mass) of KNO₃

To find the mass of KNO₃ required, we first need to find the molecular weight of KNO₃. We can calculate the molecular weight by adding up the atomic masses of the elements in the formula: potassium (K), nitrogen (N), and oxygen (O). Molecular weight of KNO₃ = Atomic weight of K + Atomic weight of N + (3 × Atomic weight of O) = 39.10 g/mol + 14.01 g/mol + (3 × 16.00 g/mol) = 101.11 g/mol

Calculate the moles of KNO₃ required

We are given the volume (150 mL) and the molarity (0.250 M) of the KNO₃solution. We can use the molarity formula to find the moles of KNO₃ that are required for the solution. Molarity (M) = \( \frac{moles of solute}{volume of solution in liters} \) 0.250 M = \( \frac{x}{0.150 L} \) To find the moles of KNO₃, we can rearrange the equation and solve for x: x = Molarity × Volume = 0.250 M × 0.150 L = 0.0375 moles of KNO₃

Calculate the mass of KNO₃ needed to prepare the solution

Now that we know the moles of KNO₃ required and the molecular weight, we can find the mass of KNO₃ needed to prepare the solution. Mass = Moles × Molecular weight = 0.0375 moles × 101.11 g/mol = 3.791625 g So, the mass of \(\mathrm{KNO}_{3}\) required to prepare 150 mL of 0.250 M \(\mathrm{KNO}_{3}\) solution is approximately 3.79 g.

Key concepts

Molarity

Molarity is an important concept in chemistry that defines the concentration of a solution. It represents how many moles of a solute are present in one liter of solution. This concept is expressed mathematically as \( M = \frac{n}{V} \), where \( M \) is the molarity, \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.
Understanding molarity is essential when making solutions of precise concentrations in lab experiments. In our exercise, we compiled the molarity to determine how much potassium nitrate (\( \mathrm{KNO}_3 \)) should be used for preparing a specified solution.
From the given molarity (0.250 M) for a solution volume of 150 mL (or 0.150 L), we calculated the moles of \( \mathrm{KNO}_3 \) required using the formula above, finding it to be 0.0375 moles. This step is fundamental as it directly impacts the mass of \( \mathrm{KNO}_3 \) that we will subsequently calculate.

Molar Mass Calculation

Molar mass, also known as molecular weight, is pivotal for converting moles to grams and vice versa. In our example, calculating the molar mass of \( \mathrm{KNO}_3 \) was a critical step. The molar mass is determined by adding the atomic masses of the constituent elements in the compound.
For \( \mathrm{KNO}_3 \), its molar mass is computed as follows:

• The atomic mass of potassium (K) is approximately 39.10 g/mol.
• The atomic mass of nitrogen (N) is about 14.01 g/mol.
• The atomic mass of oxygen (O) is roughly 16.00 g/mol.

In \( \mathrm{KNO}_3 \), we have one atom of potassium, one of nitrogen, and three of oxygen. Thus, we add their atomic masses:
\( 39.10 + 14.01 + 3 \times 16.00 = 101.11 \text{ g/mol} \).
Understanding molar mass helps translate the theoretical moles, calculated using molarity, into a practical mass measurement that can be used in lab settings.

Solution Preparation

Solution preparation involves determining how much solute is needed to prepare about a certain volume of solution at a designated concentration. This requires combining knowledge of both molarity and molar mass.
In our task, the ultimate goal was to find how many grams of \( \mathrm{KNO}_3 \) were needed. First, we determined the needed moles using the molarity formula. With a molarity of 0.250 M and a volume of 0.150 L, 0.0375 moles of \( \mathrm{KNO}_3 \) were required.
Next, we calculated the mass using the molar mass we found earlier. By multiplying the moles by the molar mass (0.0375 moles \( \times \) 101.11 g/mol), we obtained the necessary mass: approximately 3.79 g of \( \mathrm{KNO}_3 \).
Knowledge of how to correctly prepare solutions ensures precise chemical reactions, critical for successful experimentation and analysis.