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

Short Answer

Expert verified
The mass of solute is 0.494 grams.

Step by step solution

01

Understand Molarity

Molarity (M) is defined as the number of moles of solute per liter of solution. We are given a 0.0125 M solution of KMnOâ‚„, which means there are 0.0125 moles of KMnOâ‚„ in 1 liter of the solution.
02

Calculate Moles of Solute in 250 mL

The solution's volume is 250 mL, which is equivalent to 0.250 L (since 1 L = 1000 mL). To find the number of moles of solute in this volume, use the formula: number of moles = molarity × volume (L). Number of moles = 0.0125 M × 0.250 L = 0.003125 moles.
03

Find Molar Mass of KMnOâ‚„

Calculate the molar mass of KMnOâ‚„ by summing the atomic masses of each element in the compound. Potassium (K) = 39.10 g/mol, Manganese (Mn) = 54.94 g/mol, and Oxygen (O) = 16.00 g/mol (for each of the 4 oxygen atoms). Thus, the molar mass = 39.10 + 54.94 + 4(16.00) = 158.04 g/mol.
04

Calculate Mass of Solute

To find the mass of KMnO₄ in grams, use the formula: mass = number of moles × molar mass. Mass = 0.003125 moles × 158.04 g/mol = 0.493875 grams.
05

Round to Appropriate Significant Figures

Since the given molarity (0.0125 M) has three significant figures, round the final answer to three significant figures. Rounded mass = 0.494 grams.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moles of Solute
When dealing with solutions, understanding how to calculate the moles of solute is fundamental. The mole is a unit that allows chemists to count particles, akin to a dozen but on a much larger scale. Molarity, abbreviated as 'M', provides the concentration of solute within a solution and is expressed as moles per liter.
To find the moles of solute in a given volume, you apply the formula:
  • Number of moles = Molarity × Volume in Liters
This equation is essential, as it connects the molarity with the physical volume of your solution. For example, if you have a 0.0125 M solution of KMnOâ‚„ and wish to find the moles in 250 mL (or 0.250 L), simply multiply the molarity by the volume in liters to find 0.003125 moles of KMnOâ‚„.
Molar Mass Calculation
Being able to calculate the molar mass of a compound is another pivotal skill. Molar mass is the mass of one mole of a substance, in grams per mole (g/mol). For a compound like KMnOâ‚„, it's determined by summing the atomic masses of all elements in the compound's formula.
Here's how you calculate it:
  • Find each element's atomic mass from the periodic table. Potassium (K) is 39.10 g/mol, Manganese (Mn) is 54.94 g/mol, and Oxygen (O) is 16.00 g/mol.
  • For KMnOâ‚„, calculate by adding these masses: 39.10 + 54.94 + 4(16.00) = 158.04 g/mol.
This result tells us that one mole of KMnOâ‚„ weighs 158.04 grams.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. In this context, it helps us relate quantities of reactants to products using balanced chemical equations.
For solutions, stoichiometry is used to find how much solute is involved based on the moles you've calculated. Using the molarity and volume, you find moles, which then allows you to determine the mass of solute by multiplying the moles by the molar mass.
  • Mass = Moles × Molar Mass
Following this method ensures that you can convert between moles, masses, and solutions seamlessly, facilitating chemical calculations.
Significant Figures
Significant figures (sig figs) reflect the precision of your measurements. In chemistry calculations, it's important to use sig figs to communicate how accurately a number is known.
Digits in a number are deemed significant based on rules, such as:
  • Non-zero digits are always significant.
  • Any zeros between significant digits are significant.
  • Trailing zeros in a decimal portion are significant.
In our original problem, we calculated the mass of solute as 0.493875 grams. Since the molarity was given to three significant figures (0.0125), we round the final mass to 0.494 grams, following these sig fig rules. Maintaining the correct number of significant figures is crucial for accuracy in scientific communication.

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