7.4.1: Concentrations of Solutions (Problems)

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PROBLEM \(\PageIndex{1}\)

What mass of a concentrated solution of nitric acid (68.0% HNO₃ by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO₃ by mass?

Answer: 58.8 g

PROBLEM \(\PageIndex{2}\)

What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?

Answer: 375 g

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PROBLEM \(\PageIndex{3}\)

What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.

Answer: \(\mathrm{114 \;g}\)

PROBLEM \(\PageIndex{4}\)

The hardness of water (hardness count) is usually expressed in parts per million (by mass) of \(\ce{CaCO_3}\), which is equivalent to milligrams of \(\ce{CaCO_3}\) per liter of water. What is the molar concentration of Ca²⁺ ions in a water sample with a hardness count of 175 mg CaCO₃/L?

Answer: \(1.75 \times 10^{−3} M\)

PROBLEM \(\PageIndex{5}\)

A throat spray is 1.40% by mass phenol, \(\ce{C_6H_5OH}\), in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.

Answer: 0.148 M

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PROBLEM \(\PageIndex{6}\)

Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?

Answer: \(\mathrm{2.38 \times 10^{−4}\: mol}\)

PROBLEM \(\PageIndex{7}\)

What are the mole fractions of H₃PO₄ and water in a solution of 14.5 g of H₃PO₄ in 125 g of water?

Answer

\(X_\mathrm{H_3PO_4}=0.021\)

\(X_\mathrm{H_2O}=0.979\)

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PROBLEM \(\PageIndex{8}\)

What are the mole fractions of HNO₃ and water in a concentrated solution of nitric acid (68.0% HNO₃ by mass)?

Answer

\(X_\mathrm{HNO_3}=0.378\)

\(X_\mathrm{H_2O}=0.622\)

PROBLEM \(\PageIndex{9}\)

Calculate the mole fraction of each solute and solvent:

583 g of H₂SO₄ in 1.50 kg of water—the acid solution used in an automobile battery
0.86 g of NaCl in 1.00 × 10² g of water—a solution of sodium chloride for intravenous injection
46.85 g of codeine, C₁₈H₂₁NO₃, in 125.5 g of ethanol, C₂H₅OH
25 g of I₂ in 125 g of ethanol, C₂H₅OH

Answer a

\(X_\mathrm{H_2SO_4}=0.067\)

\(X_\mathrm{H_2O}=0.933\)

Answer b

\(X_\mathrm{HCl}=0.0026\)

\(X_\mathrm{H_2O}=0.9974\)

Answer c

\(X_\mathrm{codiene}=0.054\)

\(X_\mathrm{EtOH}=0.946\)

Answer d

\(X_\mathrm{I_2}=0.035\)

\(X_\mathrm{EtOH}=0.965\)

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PROBLEM \(\PageIndex{10}\)

Calculate the mole fraction of each solute and solvent:

0.710 kg of sodium carbonate (washing soda), Na₂CO₃, in 10.0 kg of water—a saturated solution at 0 °C
125 g of NH₄NO₃ in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl₂ in 125 g of dichloromethane, CH₂Cl₂
0.372 g of histamine, C₅H₉N, in 125 g of chloroform, CHCl₃

Answer a

\(X_\mathrm{Na_2CO_3}=0.0119\)

\(X_\mathrm{H_2O}=0.988\)

Answer b

\(X_\mathrm{NH_4NO_3}=0.09927\)

\(X_\mathrm{H_2O}=0.907\)

Answer c

\(X_\mathrm{Cl_2}=0.192\)

\(X_\mathrm{CH_2CI_2}=0.808\)

Answer d

\(X_\mathrm{C_5H_9N}=0.00426\)

\(X_\mathrm{CHCl_3}=0.997\)

PROBLEM \(\PageIndex{11}\)

What is the difference between a 1 M solution and a 1 m solution?

Answer: In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.

PROBLEM \(\PageIndex{12}\)

What is the molality of phosphoric acid, H₃PO₄, in a solution of 14.5 g of H₃PO₄ in 125 g of water?

Answer: 1.18 m

PROBLEM \(\PageIndex{13}\)

What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO₃ by mass)?

Answer: 33.7 m

PROBLEM \(\PageIndex{14}\)

Calculate the molality of each of the following solutions:

0.710 kg of sodium carbonate (washing soda), Na₂CO₃, in 10.0 kg of water—a saturated solution at 0°C
125 g of NH₄NO₃ in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl₂ in 125 g of dichloromethane, CH₂Cl₂
0.372 g of histamine, C₅H₉N, in 125 g of chloroform, CHCl₃

Answer: 6.70 × 10⁻¹ m

Answer: 5.67 m

Answer: 2.8 m

Answer: 0.0358 m

PROBLEM \(\PageIndex{15}\)

A 13.0% solution of K₂CO₃ by mass has a density of 1.09 g/cm³. Calculate the molality of the solution.

Answer: 1.08 m

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).
Adelaide Clark, Oregon Institute of Technology

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