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Chapter 6: Problem 12

Which of the following must be known to calculate the molarity of a salt solution (there may be more than one answer)? a. the mass of salt added b. the molar mass of the salt c. the volume of water added d. the total volume of the solution Explain.

Short Answer

Expert verified
To calculate the molarity of a salt solution, we must know the mass of salt added (a), the molar mass of the salt (b), and the total volume of the solution (d). Molarity can be determined using the formula: \( molarity = \frac{moles~of~solute}{volume~of~the~solution~in~liters}\), where the moles of solute can be calculated by dividing the mass of salt by its molar mass.

Step by step solution

01

Identify the mass of salt added and the molar mass of the salt

The mass of salt added to the solution (option a) and the molar mass of the salt (option b) are both required to calculate the moles of solute present. Make note of these values.
02

Calculate moles of solute

Divide the mass of salt added by the molar mass of the salt to find the number of moles of solute. Use the following formula: moles of solute = (mass of salt added) / (molar mass of the salt)
03

Identify the total volume of the solution

Option d states that we need the total volume of the solution. This is the final volume of the mixture of salt and water. Make note of this value.
04

Convert the total volume of the solution to liters

If the volume of the solution is not already in liters, convert the volume to liters (1 L = 1000 mL).
05

Calculate the molarity of the salt solution

Divide the moles of solute (from Step 2) by the volume of the solution in liters (from Step 4) to find the molarity of the salt solution. Use the following formula: molarity = (moles of solute) / (volume of the solution in liters) Hence, to calculate the molarity of a salt solution, we must know the mass of salt added (a), the molar mass of the salt (b), and the total volume of the solution (d).

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Key Concepts

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

Molarity of a Solution
Molarity, often denoted by the symbol M, is a unit of concentration that exemplifies the number of moles of a solute per liter of solution. This measurement is a way to express the strength or concentration of a solution in a uniform way. To calculate molarity, one must simply divide the moles of solute by the volume of the solution in liters. The formula looks like this: \[\begin{equation}Molarity (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\end{equation}\]Understanding molarity is crucial in fields such as chemistry and biology where solutions are regularly used in experiments and reactions.

For example, let's calculate the molarity of a solution where 58.5 grams of NaCl (table salt) is dissolved in enough water to make 1 liter of solution. Assuming the molar mass of NaCl is 58.45 g/mol, the calculation would entail determining the moles of NaCl using the mass and molar mass and then dividing it by the volume of the solution.
Molar Mass
Molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It corresponds to the atomic or molecular weight of a substance taken from the periodic table based on the number and type of atoms in a single molecule of the substance.

To find the molar mass of a compound, like salt (NaCl), simply add together the atomic masses of sodium (Na), around 22.99 g/mol, and chlorine (Cl), approximately 35.45 g/mol. This would give us:\[\begin{equation}Molar\;Mass\;of\;NaCl = 22.99\;g/mol\; + 35.45\;g/mol\; = 58.44\;g/mol\end{equation}\]Knowing the molar mass is essential when converting from the mass of a substance to the number of moles, which is a key step in calculating the molarity of a solution.
Moles of Solute
Moles of solute is a measure of the number of moles of a substance that is present in a given volume of solution. A mole represents Avogadro's number \(6.022 \times 10^{23}\) of entities, such as atoms, ions, or molecules. Calculating the moles of solute is essential for determining molarity and requires the mass of the solute and its molar mass.

Using the formula\[\begin{equation}Moles\;of\;solute = \frac{Mass\;of\;solute}{Molar\;mass\;of\;solute}\end{equation}\]we can establish the connection between the tangible mass of a substance and the theoretical concept of moles, which is pivotal for understanding chemical reactions and solution dynamics.
Solution Volume
Solution volume is the total space occupied by the solution, which includes both the solute and the solvent. It is typically measured in liters (L) or milliliters (mL), but for the purpose of calculating molarity, it is essential to use liters. To convert milliliters to liters, one simply divides the volume in milliliters by 1000, since 1 liter equals 1000 milliliters.

The formula for converting is:\[\begin{equation}Volume\;in\;liters = \frac{Volume\;in\;milliliters}{1000}\end{equation}\]Having the precise volume of the solution is crucial, as molarity is defined per liter of solution. Therefore, accurate volume measurements are integral to achieving correct molarity calculations.

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Most popular questions from this chapter

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

A solution was prepared by mixing \(50.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HNO_and \(100.00 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HNO \(_{3}\). Calculate the molarity of the final solution of nitric acid.

A stream flows at a rate of \(5.00 \times 10^{4}\) liters per second (L/s) upstream of a manufacturing plant. The plant discharges \(3.50 \times 10^{3} \mathrm{L} / \mathrm{s}\) of water that contains \(65.0 \mathrm{ppm}\) HCl into the stream. (See Exercise 123 for definitions.) a. Calculate the stream's total flow rate downstream from this plant. b. Calculate the concentration of HCl in ppm downstream from this plant. c. Further downstream, another manufacturing plant diverts \(1.80 \times 10^{4} \mathrm{L} / \mathrm{s}\) of water from the stream for its own use. This plant must first neutralize the acid and does so by adding lime: $$ \mathrm{CaO}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ What mass of \(\mathrm{CaO}\) is consumed in an 8.00 -h work day by this plant? d. The original stream water contained \(10.2 \mathrm{ppm} \mathrm{Ca}^{2+} .\) Although no calcium was in the waste water from the first plant, the waste water of the second plant contains \(\mathrm{Ca}^{2+}\) from the neutralization process. If \(90.0 \%\) of the water used by the second plant is returned to the stream, calculate the concentration of \(\mathrm{Ca}^{2+}\) in ppm downstream of the second plant.

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. copper(II) sulfate \((a q)+\) iron \((s) \longrightarrow\) copper \((s)\) # iron(II) sulfate \((a q)\) copper \((\text { II) sulfate }(a q) \text { # iron }(s) \Longrightarrow\) copper \((s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate 2.27 g of copper. Which equation above describes the reaction that occurred? Support your answer.

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.0200 mole of sodium phosphate in \(10.0 \mathrm{mL}\) of solution b. 0.300 mole of barium nitrate in \(600.0 \mathrm{mL}\) of solution c. \(1.00 \mathrm{g}\) of potassium chloride in \(0.500 \mathrm{L}\) of solution d. \(132 \mathrm{g}\) of ammonium sulfate in \(1.50 \mathrm{L}\) of solution
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