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Chapter 7: Problem 2

Calculate the mass of sodium chloride required to make \(125.0 \mathrm{~mL}\) of a \(0.470 \mathrm{M} \mathrm{NaCl}\) solution. If you dissolve \(5.8 \mathrm{~g}\) of \(\mathrm{NaCl}\) in water and then dilute to a total of \(100.0 \mathrm{~mL}\), what will be the molar concentration of the resulting sodium chloride solution?

Short Answer

Expert verified
To make a 0.470M NaCl solution, 7.315 g of NaCl are required for 125.0 mL. The molar concentration of the resulting solution when 5.8 g NaCl is dissolved in 100.0 mL of water is 0.992M.

Step by step solution

01

Calculate the moles of NaCl required

To determine the mass of sodium chloride (NaCl) needed to make the solution, use the formula for molarity, which is moles of solute divided by volume of solution in liters. The volume of the solution is given as 125.0 mL, which must be converted to liters by dividing by 1000. Thus, the volume is 0.125 L. Acording to the molarity formula, M = moles of solute / volume of solution in liters, \(0.470 \text{M} = \text{moles of NaCl} / 0.125 \text{L}\). Solve for moles of NaCl.
02

Convert moles of NaCl to grams

Molecular weight (MW) of NaCl is 22.99 g/mol (for Na) + 35.45 g/mol (for Cl) = 58.44 g/mol. Multiply the moles of NaCl obtained in Step 1 by the MW to find the mass in grams needed.
03

Calculate the molar concentration of the diluted NaCl

Given that 5.8 g of NaCl is dissolved in water to make 100.0 mL of solution, first calculate the moles of NaCl using its MW. Then, convert the volume from mL to L by dividing by 1000. Finally, calculate the molarity by dividing the moles of NaCl by the volume in liters.

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Key Concepts

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

Molecular Weight
Molecular weight or molecular mass is a fundamental concept in chemistry, referring to the sum of the atomic masses of the atoms in a molecule. It's typically measured in grams per mole (g/mol) and can be found on the periodic table as the atomic mass for each element.

For instance, in the case of sodium chloride (NaCl), the molecular weight can be calculated by adding the atomic masses of sodium (Na, approximately 22.99 g/mol) and chlorine (Cl, approximately 35.45 g/mol). The sum, 58.44 g/mol, represents the molecular weight of NaCl.

Understanding molecular weight is essential for converting between moles and grams, as it serves as the conversion factor in calculations.
Moles Calculation
Moles are a basic unit in chemistry that quantify the amount of a substance. One mole corresponds to Avogadro's number ((6.022 \times 10^{23})) of entities, be they atoms, ions, or molecules.

To calculate moles from mass, use the formula:\[ \text{moles} = \frac{\text{mass}}{\text{molecular weight}} \]Conversely, to find the mass when the number of moles is given, multiply the moles by the molecular weight of the substance. This step is integral in making solutions where a specific amount of substance is required.
Solution Preparation
Preparing a chemical solution requires dissolving a precise mass of solute (the substance to be dissolved) in a solvent to achieve a desired volume and concentration.

Key steps in the process include:
  • Determining the desired molarity and volume of the solution.
  • Calculating the required mass of the solute using its molecular weight and the molarity formula.
  • Dissolving the solute in a volume of solvent less than the final desired volume.
  • Diluting the solution to the final intended volume with additional solvent.

By carefully measuring and mixing these components, you can create a solution with precise specifications for your experimental needs.
Concentration of Solution
The concentration of a solution indicates the amount of solute present in a given volume of solvent and is commonly expressed in molarity (M). Molarity is defined as the number of moles of solute per liter of solution.

To calculate molarity, use the formula:\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]This formula is crucial when mixing solutions for laboratory work, ensuring that reactions proceed correctly or to achieve accurate results in various analyses.

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Most popular questions from this chapter

A sample of \(12.7\) grams of sodium sulfate (Na2SO4) is dissolved in \(672 \mathrm{~mL}\) of distilled water. a. What is the molar concentration of sodium sulfate in the solution? b. What is the concentration of sodium ion in the solution?

For each of the molecules of \(\mathrm{NH}_{3}\) and \(\mathrm{CO}_{2}\) indicate whether a molecular dipole exists. If a dipole does exist, use a dipole arrow to indicate the direction of the molecular dipole.

1\. A 1.50 mL aliquot of a \(0.177 \mathrm{M}\) solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) is diluted into \(10.0 \mathrm{~mL}\) of distilled water, to give solution A. A \(10.0 \mathrm{~mL}\) aliquot of \(\mathbf{A}\) is then diluted into \(50.0 \mathrm{~mL}\) of distilled water, to give solution \(\mathbf{B}\). Finally, \(10.0 \mathrm{~mL}\) of \(\mathbf{B}\) is diluted into \(900.0 \mathrm{~mL}\) of distilled water to give solution C. Additional distilled water is then added to \(\mathrm{C}\) to give a final volume of \(1.0000 \mathrm{~L}\). What is the final concentration of sulfuric acid in solution C? 2\. A solution was prepared by mixing \(250 \mathrm{~mL}\) of \(0.547 \mathrm{M} \mathrm{NaOH}\) with \(50.0 \mathrm{~mL}\) of \(1.62 \mathrm{M} \mathrm{NaOH}\) and then diluting to a final volume of \(1.50 \mathrm{~L}\). What is the molarity of \(\mathrm{Na}^{+}\) in this solution? To what final volume should \(75.00 \mathrm{~mL}\) of \(0.889\) M HCl(aq) be diluted to prepare \(0.800 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) ?
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