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Ch.5 - Introduction to Solutions and Aqueous Solutions
Tro - Chemistry: A Molecular Approach 6th Edition
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217Not the one you use?Change textbook
All textbooksTro 6th EditionCh.5 - Introduction to Solutions and Aqueous SolutionsProblem 33
Chapter 5, Problem 33

A chemist wants to make 3.00 L of a 0.250 M NaNO3 solution. What mass of NaNO3 (in g) should the chemist use?

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Determine the molarity formula: Molarity (M) is defined as the number of moles of solute per liter of solution. The formula is M = \(\frac{n}{V}\), where n is the number of moles and V is the volume in liters.
Rearrange the formula to solve for the number of moles of NaNO_3 needed: n = M \(\times\) V.
Substitute the given values into the equation: M = 0.250 \(\text{ M}\) and V = 3.00 \(\text{ L}\). Calculate the number of moles of NaNO_3 required.
Use the molar mass of NaNO_3 to convert moles to grams. The molar mass of NaNO_3 is approximately 85.00 \(\text{ g/mol}\).
Multiply the number of moles by the molar mass to find the mass of NaNO_3 needed in grams.

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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, the desired molarity of the NaNO3 solution is 0.250 M, indicating that there are 0.250 moles of NaNO3 in every liter of the solution.
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Moles and Mass Relationship

The relationship between moles and mass is governed by the formula: mass = moles × molar mass. To find the mass of NaNO3 needed, the chemist must first calculate the total number of moles required for the 3.00 L solution and then multiply that by the molar mass of NaNO3, which is approximately 85.00 g/mol.
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Dilution and Solution Preparation

Preparing a solution involves accurately measuring and mixing a solute with a solvent to achieve a desired concentration. In this case, the chemist needs to ensure that the calculated mass of NaNO3 is properly dissolved in enough water to make a total volume of 3.00 L, ensuring the final concentration is 0.250 M.
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