A laboratory procedure calls for making 400.0 mL of a 1.1 M NaNO 3 solution. What mass of NaNO 3 (in g) is needed?

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Ch.5 - Introduction to Solutions and Aqueous Solutions

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Tro 5th Edition

Ch.5 - Introduction to Solutions and Aqueous Solutions

Problem 27

Chapter 5, Problem 27

A laboratory procedure calls for making 400.0 mL of a 1.1 M NaNO₃ solution. What mass of NaNO₃ (in g) is needed?

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Calculate the number of moles of NaNO_3 needed using the formula: \( \text{moles} = \text{molarity} \times \text{volume in liters} \). Convert 400.0 mL to liters by dividing by 1000.

Use the molarity (1.1 M) and the volume in liters from Step 1 to find the moles of NaNO_3.

Determine the molar mass of NaNO_3 by adding the atomic masses of Na, N, and O (Na: 22.99 g/mol, N: 14.01 g/mol, O: 16.00 g/mol).

Calculate the mass of NaNO_3 needed using the formula: \( \text{mass} = \text{moles} \times \text{molar mass} \).

Use the moles from Step 2 and the molar mass from Step 3 to find the mass of NaNO_3 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, a 1.1 M NaNO3 solution means there are 1.1 moles of sodium nitrate in every liter of the solution.

Calculating Moles from Volume and Molarity

To find the number of moles of solute needed for a specific volume of solution, you can use the formula: moles = molarity × volume (in liters). For a 400.0 mL solution, you must convert the volume to liters (0.400 L) before applying this formula to determine the moles of NaNO3 required.

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 NaNO3, the molar mass can be calculated by summing the atomic masses of sodium (Na), nitrogen (N), and three oxygen (O) atoms. This value is essential for converting moles of NaNO3 into grams.