Problem 41 - Chemistry and Chemical Reactivity

Question

What is the mass of solute, in grams, in \(250 . \mathrm{mL}\) of a 0.0125 M solution of \(\mathrm{KMnO}_{4} ?\)

Step-by-Step Solution

Answer

The mass of solute is 0.494 g.

Step 1: Identify the Components in the Problem

We need to find the mass of solute, given that we have 250 mL of a 0.0125 M solution. The solute in this solution is KMnO₄.

Step 2: Convert Volume from mL to L

Since molarity is expressed in moles per liter, we need to convert 250 mL to liters. 250 mL = 0.250 L.

Step 3: Calculate Moles of Solute

Use the molarity formula: \[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume in liters}} \]Rearrange to find moles of solute: \[ \text{Moles of solute} = \text{Molarity} \times \text{Volume in liters} \]\[ \text{Moles of solute} = 0.0125 \, M \times 0.250 \, L = 0.003125 \, ext{moles} \]

Step 4: Determine Molar Mass of KMnO₄

Add the atomic masses of all atoms in KMnO₄: - K: 39.1 g/mol - Mn: 54.9 g/mol - O (4 atoms): 4 × 16.0 g/mol = 64.0 g/mol Thus, molar mass of KMnO₄ = 39.1 + 54.9 + 64.0 = 158.0 g/mol.

Step 5: Calculate Mass of Solute

Multiply the number of moles by the molar mass to get the mass of the solute:\[ \text{Mass of solute} = \text{Moles of solute} \times \text{Molar mass} \]\[ \text{Mass of solute} = 0.003125 \, ext{moles} \times 158.0 \, \text{g/mol} = 0.49375 \, \text{g} \].

Step 6: Round Off to Appropriate Sig Figs

The given molarity (0.0125 M) and volume (250 mL) both have three significant figures. Therefore, round the mass to three significant figures: 0.494 g.

Key Concepts

Understanding Molarity

Molarity is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution, denoted by the symbol M. Understanding molarity is crucial in chemistry as it allows us to determine how much solute is dissolved in a given volume of solvent.

To calculate molarity, you can use the formula:

• Molarity (M) = Moles of solute / Volume of solution in liters

This formula is essential for reactions and calculations involving solutions, where knowing the concentration helps in predicting how substances will react with each other. In the given exercise, the molarity of the KMnO₄ solution is 0.0125 M, indicating that 0.0125 moles of KMnO₄ are present in each liter of solution.

Conversion of Units in Chemistry

Converting units is an essential skill in chemistry, as it ensures that calculations are accurate and consistent with the units used. Since molarity involves liters, we often need to convert volumes given in milliliters (mL) to liters (L).

To convert milliliters to liters, remember that 1 liter = 1000 milliliters. Thus, you can convert by dividing the volume in milliliters by 1000:

• Volume in liters = Volume in mL / 1000

In the exercise, 250 mL of solution is converted to 0.250 L. Being comfortable with these conversions is vital as you'll often encounter different units in experiments and problems. Precise conversions are needed to ensure the accuracy of your calculations.

Significance of Significant Figures

Significant figures are the digits in a measurement that contribute to its precision. When performing calculations, it's important to use the correct number of significant figures, as they represent the accuracy of your measurements.

In chemistry, when multiplying or dividing, the number of significant figures in the result should match the least number of significant figures in any of the values used. For addition or subtraction, align to the least number of decimal places.

• The step-by-step solution rounds the calculated mass to three significant figures because the given molarity (0.0125 M) and volume (250 mL) both have three significant figures.

Using significant figures properly helps communicate the precision of your results and is a critical component of scientific practice.

Basics of Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's based on the law of conservation of mass and uses balanced chemical equations to relate amounts in moles of different substances.

Understanding stoichiometry is essential for predicting reaction yields, determining limiting reagents, and converting between moles and grams using molar masses. In the exercise, stoichiometry helps calculate the mass of solute using the formula:

• Mass of solute = Moles of solute × Molar mass

Given that the molar mass of KMnO₄ is calculated as 158.0 g/mol, multiplying it by the moles of solute (0.003125 moles) gives the mass of KMnO₄. Mastery of stoichiometry enables accurate predictions in lab settings and ensures effective communication of chemical phenomena.