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Ch.5 - Introduction to Solutions and Aqueous Solutions
Chapter 5, Problem 31

A laboratory procedure calls for making 100.0 mL of a 1.30 M K2SO4 solution. What mass of K2SO4 (in g) is needed?

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Determine the number of moles of K_2SO_4 needed using the formula: \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \).
Convert the volume of the solution from milliliters to liters: \( 100.0 \text{ mL} = 0.1000 \text{ L} \).
Calculate the moles of K_2SO_4 required using the molarity and volume: \( \text{moles of K}_2\text{SO}_4 = 1.30 \text{ M} \times 0.1000 \text{ L} \).
Find the molar mass of K_2SO_4 by adding the atomic masses of all atoms in the formula: \( 2 \times \text{K} + 1 \times \text{S} + 4 \times \text{O} \).
Calculate the mass of K_2SO_4 needed using the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity (M)

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in moles per liter (mol/L). In this question, a 1.30 M solution indicates that there are 1.30 moles of K2SO4 in every liter of the solution, which is crucial for calculating the mass needed for the desired volume.
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Molar Mass

The molar mass of a compound is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). For K2SO4, the molar mass can be calculated by summing the atomic masses of its constituent elements: potassium (K), sulfur (S), and oxygen (O). This value is essential for converting moles of K2SO4 into grams.
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Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In this context, it allows us to use the molarity and volume of the solution to determine the number of moles of K2SO4 required, which can then be converted to mass using the molar mass.
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