(a) Determine the mass of anhydrous copper(II) sulfate that must be used to prepare \(250 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution. (b) Determine the mass of \(\mathrm{CuSO}_{4}-5 \mathrm{H}_{2} \mathrm{O}\) that must be used to prepare \(250 \mathrm{ml}\). of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}(\mathrm{aq})\) solution.

Short Answer

Expert verified
To prepare a 0.20 M CuSO4 solution, use 7.98 g of anhydrous CuSO4 and 12.48 g of CuSO4*5H2O for a volume of 250 mL.

Step by step solution

01

Calculate moles of CuSO4

Use the molarity formula, M = moles / volume (L), to find the moles of CuSO4. Given M = 0.20 M and V = 0.250 L, moles = M x V.
02

Convert moles to mass for anhydrous CuSO4

Use the molar mass of anhydrous CuSO4 (159.61 g/mol) to convert the moles from step 1 to mass. Mass = moles x molar mass.
03

Calculate the mass of hydrated CuSO4*5H2O needed

Use the molar mass of CuSO4*5H2O (249.68 g/mol) and the moles of CuSO4 from step 1 to determine the mass needed. Mass = moles x molar mass of CuSO4*5H2O.

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Key Concepts

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

Solution Preparation
Preparing chemical solutions is a fundamental skill in chemistry that requires precision and understanding of materials and measurements. A solution consists of a solute, which in this case is copper(II) sulfate, dissolved in a solvent, most often water. The concentration of solutions is often described in molarity (M), which is moles of solute per liter of solution.

To prepare a specific molarity, one must first calculate the number of moles of solute needed, using the molarity equation: \( M = \frac{ \text{moles of solute} }{ \text{volume of solution in liters} } \). With the number of moles and the molar mass of the solute, we can then calculate the mass of the solute required. It's vital to then accurately weigh out this mass and dissolve it completely in a volume of solvent to achieve the desired total volume and concentration.
Molar Mass
The molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It's fundamental in converting between the mass of a substance and the number of moles. Each element's molar mass can be found in the periodic table as the atomic weight.

For compounds, the molar mass is the sum of the atomic weights of each constituent element, multiplied by the number of atoms of that element in the molecular formula. For example, in anhydrous copper(II) sulfate (CuSO4), one would add up the molar masses of one copper atom, one sulfur atom, and four oxygen atoms. Molar mass calculations are essential for the steps involved in solution preparation, as they link the mass of a substance to its mole quantity.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the context of solution preparation, stoichiometry is used to determine the amount of solute needed to attain a specific concentration.

Stoichiometric Calculations in Solution Preparation

For example, one can use stoichiometry to convert moles of a substance to mass by employing the molar mass. This allows us to calculate how much of a compound to weigh out in order to prepare a solution of a given concentration. Chemistry relies on stoichiometry to ensure that reactions have the correct proportion of reactants, which is crucial for both laboratory work and industrial processes.
Chemistry Education
Effective chemistry education is critical as it lays the foundation for understanding complex concepts and techniques within the scientific world. Mastering the skills needed for solution preparation, understanding molar mass, and applicating stoichiometry goes beyond memory; it develops critical thinking and problem-solving abilities.

Integrating real-world examples, laboratory experiences, and interactive learning aids students in the absorption of fundamental concepts like those discussed in this exercise. The goal is to break down complex ideas into easy-to-grasp chunks that enable students to apply these concepts confidently in practical situations, supporting their journey in the fascinating world of chemistry.

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Most popular questions from this chapter

(a) A 12.56-mL sample of \(1.345 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{~mL}\). What is the molar concentration of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the diluted solution? (b) A \(25.00-\mathrm{mL}\) sample of \(0.366 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) is drawn from a reagent bottle with a pipet. The sample is transferred to a \(125.00-\mathrm{mL}\) volumetric flask and diluted to the mark with water. What is the molar concentration of the dilute hydrochloric acid solution?

A chemist prepared an aqueous solution by mixing \(2.50 \mathrm{~g}\) of ammonium phosphate trihydrate, \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) and \(1.50 \mathrm{~g}\) of potassium phosphate, \(\mathrm{K}_{3} \mathrm{PO}_{4}\), with \(500 \mathrm{~g}\) of water. (a) Determine the number of moles of formula units of each compound that was measured. (b) How many moles of \(\mathrm{PO}_{4}{ }^{3-}\) are present in solution? (c) Calculate the mass of phosphate ions present in the solution. (d) What is the total mass of the water present in the solution?

Nitric acid is purchased from chemical suppliers as a solution that is \(70 \% \mathrm{HNO}_{3}\) by mass. What mass (in grams) of a \(70 \% \mathrm{HNO}_{3}(\mathrm{aq})\) solution is required to prepare \(250 \mathrm{~g}\) of a \(2.0 \mathrm{~m} \mathrm{HNO}_{3}\) (aq) solution? The density of \(70 \% \mathrm{HNO}_{3}(\mathrm{aq})\) is \(1.42 \mathrm{~g} \cdot \mathrm{cm}^{-1}\).

Calculate the mole fraction of each component in the following solutions: (a) \(2.5 .0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\); (b) \(25.0 \mathrm{~g}\) of water and \(50.0 \mathrm{~g}\) of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\); (c) a glucowe solution that is \(0.10 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\).

To prepare a very dilute solution, it is advisable to perform successive dilutions of a single prepared reagent solution, rather than to wcigh a very small mass or to measure a very small volume of stock chemical. A solution was prepared by transferring \(0.661 \mathrm{~g}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to a \(250.0\)-mL volumetric flask and adding water to the mark. A \(1.000-\mathrm{mL}\). sample of this solution. was transferred to a 500 -mL. volumetric flask and diluted to the mark with water. Then \(10.0 \mathrm{~mL}\) of the diluted solution was transferred to a 250 -mL. flask and diluted to the mark with water. (a) What is the final concentration of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{\text {, in solution? (b) What }}\) mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is in this final solution? (The answer to the last question gives the amount that would have had to have been weighed out if the solution had been prepared directly.)

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