Module 2.3 Concentration and Molarity

A solution is prepared by dissolving 1.25 g of Ca(OH)₂ in enough water to produce 1.00L of solution.

What is the concentration of the hydroxide ions in the solution?

• 3.37 × 10^-2 g L^-1

• 1.69 × 10^-2 g L^-1

• 1.69 × 10^-2 mol L^-1

• 3.37 × 10^-2 mol L^-1

Answer explanation

To determine the concentration of hydroxide ions in a solution prepared by dissolving 1.25 grams of Ca(OH)₂ in 1.00 liter of water, first calculate the molar mass of Ca(OH)₂: 40.1 g/mol (Ca) + 32.0 g/mol (O) + 2.0 g/mol (H) = 74.1 g/mol. The number of moles in 1.25 grams is 1.25 g / 74.1 g/mol ≈ 0.0169 mol. The concentration of Ca(OH)₂ is 0.0169 mol / 1.00 L = 0.0169 mol/L. Since each Ca(OH)₂ molecule produces 2 hydroxide ions, the hydroxide ion concentration is 0.0169 mol/L × 2 = 0.0338 mol/L. Thus, the concentration of hydroxide ions is approximately 3.37 × 10⁻² mol/L.

The concentration of a solution of ammonia (NH₃) is 2.50% (w/v).

What is the molar concentration produced by diluting 25.0 mL of this solution with 250 mL of water?

• 0.13 mol L^-1

• 0.147 mol L^-1

• 1.47 mol L^-1

• 1.30 mol L^-1

Answer explanation

To determine the molar concentration of ammonia (NH₃) after diluting a 2.50% (w/v) solution, first note that 2.50% (w/v) corresponds to 2.50 grams of NH₃ per 100 mL of solution. In 25.0 mL of this solution, there are 2.50 g / 100 mL × 25.0 mL = 0.625 g of NH₃. With a molar mass of 17.0 g/mol for NH₃, this amounts to 0.625 g / 17.0 g/mol ≈ 0.0368 mol. After diluting to a total volume of 275 mL (0.275 L), the molar concentration is 0.0368 mol / 0.275 L ≈ 0.134 mol/L. Therefore, the molar concentration of the diluted solution is approximately 0.13 mol/L.

A saline solution is made by dissolving pure sodium chloride (NaCl) in distilled water. Two of its uses are to help remove contact lenses and to relieve a condition called "dry eyes". A 20.0 mL dispenser of eye drops contains 1.5 milligrams of sodium chloride.

What is the concentration of sodium chloride solution in the dispenser?

• 1.3 × 10^-3 mol L^-1

• 2.6 × 10^-2 mol L^-1

• 3.9 × 10^-2 mol L^-1

• 5.1 × 10^-1 mol L^-1

Answer explanation

To determine the concentration of sodium chloride (NaCl) in a 20.0 mL dispenser containing 1.5 milligrams of NaCl, first convert 1.5 milligrams to grams (0.0015 g). The molar mass of NaCl is 58.44 g/mol, so the number of moles in 0.0015 grams is 0.0015 g / 58.44 g/mol ≈ 2.57×10−5 mol. Converting 20.0 mL to liters gives 0.020 L. The concentration is then 2.57×10−5 mol / 0.020 L ≈ 1.29×10−3 mol/L. Therefore, the concentration of the NaCl solution in the dispenser is approximately 1.3 × 10⁻³ mol/L.

A sample of 5.3 grams of sodium carbonate (Na₂CO₃) is dissolved in distilled water and the solution is made up to 100.0 mL. A sample of 20.0 mL of this solution is placed into a volumetric flask and distilled water is added until a volume of 200.0 mL is achieved.

What is the concentration of sodium ions (Na⁺) in the volumetric flask?

• 0.050 mol L^-1

• 0.10 mol L^-1

• 0.20 mol L^-1

• 0.50 mol L^-1

Answer explanation

To find the concentration of sodium ions (Na⁺) in the final solution, start by calculating the molar concentration of sodium carbonate (Na₂CO₃). The molar mass of Na₂CO₃ is 106 g/mol, so 5.3 grams of Na₂CO₃ corresponds to approximately 0.050 moles. This amount dissolved in 100.0 mL results in an initial concentration of 0.50 mol/L. When 20.0 mL of this solution is diluted to 200.0 mL, the concentration becomes 0.050 mol/L. Since Na₂CO₃ dissociates into 2 Na⁺ ions per molecule, the final concentration of Na⁺ ions is 0.10 mol/L. Thus, the concentration of sodium ions in the volumetric flask is 0.10 mol/L.

How much water would need to be added to 650 mL of a 2.70 mol L^-1 HCl solution to make a 0.500 mol L^-1 solution?

• 3.51 L

• 1.43 L

• 2.08 L

• 2.86 L

Answer explanation

To dilute 650 mL of a 2.70 mol/L HCl solution to a final concentration of 0.500 mol/L, you need to determine the total volume required using the dilution formula. The final volume should be 3.51 litres to achieve the desired concentration. Therefore, you need to add water to the initial 650 mL of solution to reach this total volume. Subtracting the initial volume from the final volume gives you the amount of water needed, which is 2.86 litres.

Which of the following best describes a standard solution?

• A solution that has known concentration

• A solution that was made using a volumetric flask

• A solution that has a concentration of 1.0 mol L^-1 at 25^oC and 100 kPa

• A solution that was made at room temperature and pressure

Answer explanation

A standard solution is best described as a solution with a known concentration. This definition is crucial because standard solutions are prepared to have a precise concentration for accurate analytical work. The other options are incorrect because the method of preparation or specific conditions (like using a volumetric flask, having a concentration of 1.0 mol L-1 at specific conditions, or being made at room temperature) do not inherently guarantee a known concentration.

Which of the following gives the highest concentration of chloride ions in solution?

• 1.0 moles of AlCl₃ dissolved in 500 mL water

• 1.0 moles of MgCl₂ dissolved in 250 mL water

• 0.5 moles of LiCl dissolved in 500 mL water

• 0.5 moles of FeCl₃ dissolved in 250 mL water

Answer explanation

To determine the highest concentration of chloride ions, calculate the concentration for each option. 1.0 moles of AlCl3 in 500 mL yields a chloride ion concentration of 6 M because AlCl3 dissociates into three chloride ions per formula unit. 1.0 moles of MgCl2 in 250 mL provides the highest concentration of 8 M, as MgCl2 dissociates into two chloride ions per formula unit and is in a smaller volume. 0.5 moles of LiCl in 500 mL results in a chloride concentration of 1 M, with LiCl dissociating into one chloride ion per formula unit. Lastly, 0.5 moles of FeCl3 in 250 mL gives a chloride concentration of 6 M, as FeCl3 dissociates into three chloride ions per formula unit. Thus, MgCl2 produces the highest chloride ion concentration.

A sample of 10 g of solid zinc metal, Zn, is placed in a flask with 200 mL of 2.0 mol L^-1 AgNO₃. The reaction that occurs is:

Zn(s) + 2AgNO₃(aq) → 2Ag(s) + Zn(NO₃)₂(aq)

The flask is set aside until no further reaction takes place.

What mass of zinc remains in the flask?

• 0 g

• 3.1 g

• 6.2 g

• 9.6 g