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

Short Answer

Expert verified
The mass of solute is 0.494 g.

Step by step solution

01

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 KMnOâ‚„.
02

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.
03

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: \[ \text{Moles of solute} = \text{Molarity} \times \text{Volume in liters} \]\[ \text{Moles of solute} = 0.0125 \, M \times 0.250 \, L = 0.003125 \, ext{moles} \]
04

Determine Molar Mass of KMnOâ‚„

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.
05

Calculate Mass of Solute

Multiply the number of moles by the molar mass to get the mass of the solute:\[ \text{Mass of solute} = \text{Moles of solute} \times \text{Molar mass} \]\[ \text{Mass of solute} = 0.003125 \, ext{moles} \times 158.0 \, \text{g/mol} = 0.49375 \, \text{g} \].
06

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.

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Key Concepts

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

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 KMnOâ‚„ solution is 0.0125 M, indicating that 0.0125 moles of KMnOâ‚„ 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 KMnOâ‚„ is calculated as 158.0 g/mol, multiplying it by the moles of solute (0.003125 moles) gives the mass of KMnOâ‚„. Mastery of stoichiometry enables accurate predictions in lab settings and ensures effective communication of chemical phenomena.

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Most popular questions from this chapter

A \(4.000-\mathrm{g}\) sample containing \(\mathrm{KCl}\) and \(\mathrm{KClO}_{4}\) was dissolved in sufficient water to give \(250.00 \mathrm{mL}\) of solution. A 50.00 -mL portion of the solution required \(41.00 \mathrm{mL}\) of \(0.0750 \mathrm{M} \mathrm{AgNO}_{3}\) in a Mohr titration (page 209 ). Next, a 25.00 -mL portion of the original solution was treated with \(\mathrm{V}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) to reduce the perchlorate ion to chloride, $$\begin{aligned} 8 \mathrm{V}^{3+}(\mathrm{aq})+\mathrm{ClO}_{4}^{-}(\mathrm{aq})+12 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow & \\ \mathrm{Cl}^{-}(\mathrm{aq})+8 \mathrm{VO}^{2+}(\mathrm{aq}) &+8 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \end{aligned}$$ and the resulting solution was titrated with \(\mathrm{AgNO}_{3} .\) This titration required \(38.12 \mathrm{mL}\) of \(0.0750 \mathrm{M} \mathrm{AgNO}_{3} .\) What is the mass percent of KCl and \(\mathrm{KClO}_{4}\) in the mixture?

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{O}_{z}\) You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol},\) what is the molecular formula?

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion. (a) \(0.12 \mathrm{M} \mathrm{BaCl}_{2}\) (b) \(0.0125 \mathrm{M} \mathrm{CuSO}_{4}\) (c) \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

If \(4.00 \mathrm{mL}\) of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) is diluted to \(10.0 \mathrm{mL}\) with pure water, what is the molar concentration of copper(II) sulfate in the diluted solution?

The reaction of \(750 .\) g each of \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\) was found to produce \(562 \text { g of } \mathrm{NO} \text { (see pages } 177-179)\). $$4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell)$$ (a) What mass of water is produced by this reaction? (b) What mass of \(\mathrm{O}_{2}\) is required to consume \(750 . \mathrm{g}\) of \(\mathrm{NH}_{3} ?\)

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