Problem 12 - General Chemistry: Principles and Modern Applications

Question

You are asked to prepare \(125.0 \mathrm{mL}\) of \(0.0321 \mathrm{M} \mathrm{AgNO}_{3}\) How many grams would you need of a sample known to be \(99.81 \% \mathrm{AgNO}_{3}\) by mass?

Step-by-Step Solution

Answer

You would need approximately 0.6823 grams of the \(99.81\% \,AgNO3\) sample.

Step 1: Understand the Question and Identify the Given Information

The question provides the following information:1. Volume (V) of the AgNO3 solution to be prepared = 125.0 mL or 0.125 L (Since one liter (L) is equivalent to 1000 mL).2. Molarity (M) of the AgNO3 solution to prepare = 0.0321 M.3. Purity of the AgNO3 sample = 99.81%.

Step 2: Calculate the Moles of AgNO3 Needed

Using the molarity formula \(M = \frac{n}{V}\), where n represents number of moles and V is the volume, we can calculate the moles of AgNO3 needed for the solution. Rearranging the formula gives \(n = M \times V\), so \(n = 0.0321 M \times 0.125 L = 0.0040125 mol\).

Step 3: Convert Moles to Grams

Using the molar mass of AgNO3 (169.87 g/mol), calculated from the sum of the molar masses of its constituent elements, the moles of AgNO3 are converted to grams with the formula \(mass = n \times molar \,mass\). This gives \(0.0040125 mol \times 169.87 g/mol = 0.6813 g\). However, this represents the mass needed if the sample was 100% pure.

Step 4: Adjusting for the Purity of the Sample

Since the sample is only 99.81% pure, the mass calculated above represents only 99.81% of the total sample we need. Therefore, the total sample needed can be calculated as \(\frac{0.6813 g}{0.9981} = 0.6823 g\), adjusting for the purity of the sample.

Key Concepts

Molarity

Molarity is a concept in chemistry used to express the concentration of a solution. It tells us how many moles of a solute are present in one liter of solution. Molarity is denoted by the symbol \( M \) and is calculated using the formula: \[ M = \frac{n}{V} \] where \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.
This measure is crucial because it helps determine the proportions necessary to achieve a desired reaction outcome or concentration.
Understanding molarity can also assist in scaling reactions to different volumes and calculating reactant amounts needed in chemical reactions.

• It enables scientists to accurately create solutions with specific concentrations.
• It simplifies the comparison of concentrations between different solutions.
• It plays a critical role in titration and other processes where precise chemical reactions are planned.

Solution Preparation

Solution preparation in chemistry involves mixing a solute with a solvent to get a solution of desired concentration and volume. In our example, the task was to prepare a certain volume of a silver nitrate (AgNO_{3}) solution with a specific molarity.
To make a solution, follow these steps:

• Calculate the required moles of solute using the molarity formula: \( n = M \times V \), where \( M \) is the desired molarity and \( V \) is the volume in liters.
• Convert these moles to grams using the molar mass formula: \( mass = n \times molar \, mass \)
• Add the calculated mass of solute to a container and dissolve it in a solvent, usually water, to achieve the desired volume.

This method ensures that the concentration is accurate for experiments or reactions where precise chemical interactions are necessary.
In our exercise, once the moles were known, they were multiplied by the molar mass of AgNO_{3} to get the required mass needed for solution preparation.

Purity Calculation

Purity calculation is essential when dealing with chemical compounds because many samples are not 100% pure. The amount of a compound that effectively contributes to the chemistry problem is determined by its purity.
When preparing a solution with a sample that isn't entirely pure, one must adjust the total mass of the sample used. The formula for adjusting the mass of a less pure sample is:

• Calculate the unadjusted mass: \( mass = n \times molar \, mass \).
• Adjust for purity: \[ total \: sample \: mass = \frac{calculated \: mass}{purity \: fraction} \]

In the exercise, the AgNO_3 sample was 99.81% pure, meaning only 99.81% contributed to the molarity. Using the adjustment formula, we ensure the effective quantity of AgNO3 in our solution meets the specified molarity. This precision is vital in experiments where every interaction counts.