Unit 3: Amount of Substance - Solutions

Mastering the language of solutions: concentration, preparation, and dilution.

3.10 Concentration Terms

Concentration describes how much solute is dissolved in a certain amount of solvent or solution. Different units are used to express concentration depending on the context.

Molarity (M)

Molarity is the most common unit of concentration in chemistry. It measures the number of moles of solute dissolved in a specific volume of the solution. Because volume can change with temperature, molarity is temperature-dependent.

Formula: $$\text{Molarity (M)} = \frac{\text{Moles of Solute (mol)}}{\text{Volume of Solution (L)}}$$

Mass Concentration

Mass concentration expresses the mass of a solute (in grams) dissolved in a specific volume of the solution (in liters or cubic decimeters). This unit is straightforward as it relates directly to the weight of the solute.

Formula: $$\text{Mass Concentration (g/L)} = \frac{\text{Mass of Solute (g)}}{\text{Volume of Solution (L)}}$$

Molality (m)

Molality measures the number of moles of solute dissolved in a specific mass of the solvent (in kilograms). Unlike molarity, molality is independent of temperature changes because it is based on mass, not volume.

Formula: $$\text{Molality (m)} = \frac{\text{Moles of Solute (mol)}}{\text{Mass of Solvent (kg)}}$$

Normality (N)

Normality is a concentration unit that considers the number of equivalents of a solute per liter of solution. An "equivalent" is the amount of a substance that can react to supply one mole of reactive species (like H⁺ ions in an acid or electrons in a redox reaction).

Formula: $$\text{Normality (N)} = \frac{\text{Number of Equivalents}}{\text{Volume of Solution (L)}}$$ It can also be calculated as: Normality = Molarity × n, where n is the number of equivalents per mole of the solute (e.g., for H₂SO₄, n = 2).

Percent Weight (% w/w)

Percent by weight (or percent by mass) describes the mass of the solute as a percentage of the total mass of the solution. It is a simple and direct way to express concentration without needing to calculate moles.

Formula: $$\text{Percent Weight (\% w/w)} = \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100\%$$ The mass of the solution is the mass of the solute plus the mass of the solvent.
Solved Examples:
  1. Molarity: Calculate the molarity of a solution prepared by dissolving 29.22 g of NaCl in enough water to make 500 mL of solution. (Molar mass of NaCl = 58.44 g/mol)
    Solution:
    • Number of moles = $29.22 \, \text{g} \div 58.44 \, \text{g/mol} = 0.5 \, \text{mol}$
    • Volume = $500 \, \text{mL} = 0.5 \, \text{L}$
    • Molarity (M) = $\frac{0.5 \, \text{mol}}{0.5 \, \text{L}} = 1.0 \, \text{M}$
  2. Mass Concentration: What is the mass concentration of a solution containing 10 g of glucose in 2 L of solution?
    Solution: Mass Concentration = $\frac{10 \, \text{g}}{2 \, \text{L}} = 5 \, \text{g/L}$
  3. Molality: 30 g of NaOH is dissolved in 250 g of water. Calculate the molality of the solution. (Molar mass of NaOH = 40.0 g/mol)
    Solution:
    • Number of moles = $30 \, \text{g} \div 40.0 \, \text{g/mol} = 0.75 \, \text{mol}$
    • Mass of solvent = $250 \, \text{g} = 0.25 \, \text{kg}$
    • Molality (m) = $\frac{0.75 \, \text{mol}}{0.25 \, \text{kg}} = 3.0 \, \text{m}$
  4. Normality: Find the normality of a 0.5 M solution of H₂SO₄.
    Solution: H₂SO₄ provides 2 equivalents (H⁺ ions) per mole.
    • Normality (N) = Molarity × Number of equivalents = $0.5 \, \text{M} \times 2 = 1.0 \, \text{N}$
  5. Percent Weight (% w/w): If 20 g of sugar is dissolved in 80 g of water, what is the percent weight of the sugar in the solution?
    Solution:
    • Mass of solution = $20 \, \text{g} + 80 \, \text{g} = 100 \, \text{g}$
    • Percent Weight = $\left(\frac{20 \, \text{g}}{100 \, \text{g}}\right) \times 100\% = 20\%$ w/w

3.11 Preparation of Standard Solutions

A standard solution is a solution containing a precisely known concentration of an element or a substance. It is prepared by dissolving a known mass of solute in a specific volume of solvent.

The process involves:

  1. Calculating the required mass of the solute.
  2. Accurately weighing the solute.
  3. Dissolving the solute in a small amount of solvent in a beaker.
  4. Transferring the solution into a volumetric flask.
  5. Adding solvent up to the calibration mark.

Key Formula: Mass of solute (g) = Molarity (mol/L) × Volume (L) × Molar Mass (g/mol)
Solved Examples:
  1. How would you prepare 250 mL of a 0.1 M NaOH solution? (Molar mass = 40.0 g/mol)
    Solution: Mass = $0.1 \, \text{mol/L} \times 0.250 \, \text{L} \times 40.0 \, \text{g/mol} = 1.0 \, \text{g}$. Weigh 1.0 g of NaOH and dissolve it in water to make a final volume of 250 mL.
  2. What mass of K₂Cr₂O₇ is needed to prepare 500 mL of a 0.02 M solution? (Molar mass = 294.2 g/mol)
    Solution: Mass = $0.02 \, \text{mol/L} \times 0.500 \, \text{L} \times 294.2 \, \text{g/mol} = 2.942 \, \text{g}$.
  3. Describe the preparation of 100 mL of 0.5 M CuSO₄ solution. (Molar mass = 159.6 g/mol)
    Solution: Mass = $0.5 \, \text{mol/L} \times 0.100 \, \text{L} \times 159.6 \, \text{g/mol} = 7.98 \, \text{g}$. Weigh 7.98 g of CuSO₄ and dissolve it in water to a final volume of 100 mL.
  4. Calculate the mass of glucose (C₆H₁₂O₆) required to make 1 L of a 0.2 M solution. (Molar mass = 180.16 g/mol)
    Solution: Mass = $0.2 \, \text{mol/L} \times 1.0 \, \text{L} \times 180.16 \, \text{g/mol} = 36.032 \, \text{g}$.
  5. To prepare 2.0 L of a 1.5 M HCl solution, what mass of HCl gas is needed? (Molar mass = 36.46 g/mol)
    Solution: Mass = $1.5 \, \text{mol/L} \times 2.0 \, \text{L} \times 36.46 \, \text{g/mol} = 109.38 \, \text{g}$.

3.12 Dilution of Standard Solutions

Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent.

When a solution is diluted, the amount of solute remains constant. This principle gives rise to the dilution formula, sometimes referred to as the dilution factor calculation.

Key Formula: $M_1V_1 = M_2V_2$
Where:
  • $M_1$ = Initial Molarity
  • $V_1$ = Initial Volume
  • $M_2$ = Final Molarity
  • $V_2$ = Final Volume
Solved Examples:
  1. If you have 100 mL of a 2.0 M HCl solution, and you dilute it to 1.0 L, what is the new concentration?
    Solution: $M_2 = \frac{M_1V_1}{V_2} = \frac{(2.0 \, \text{M})(0.1 \, \text{L})}{1.0 \, \text{L}} = 0.2 \, \text{M}$.
  2. What volume of 12 M HCl is needed to prepare 250 mL of a 1.5 M HCl solution?
    Solution: $V_1 = \frac{M_2V_2}{M_1} = \frac{(1.5 \, \text{M})(250 \, \text{mL})}{12 \, \text{M}} = 31.25 \, \text{mL}$.
  3. To what final volume must 50 mL of a 5.0 M NaOH solution be diluted to obtain a 0.5 M solution?
    Solution: $V_2 = \frac{M_1V_1}{M_2} = \frac{(5.0 \, \text{M})(50 \, \text{mL})}{0.5 \, \text{M}} = 500 \, \text{mL}$.
  4. How much water must be added to 200 mL of a 0.75 M solution to make it 0.25 M?
    Solution: Final volume $V_2 = \frac{(0.75 \, \text{M})(200 \, \text{mL})}{0.25 \, \text{M}} = 600 \, \text{mL}$. Water added = $600 \, \text{mL} - 200 \, \text{mL} = 400 \, \text{mL}$.
  5. A student takes 25 mL of a 4.0 M stock solution and dilutes it to a final volume of 200 mL. What is the final molarity?
    Solution: $M_2 = \frac{(4.0 \, \text{M})(25 \, \text{mL})}{200 \, \text{mL}} = 0.5 \, \text{M}$.

End of Topic Questions

Answer: Moles of solute per liter of solution.

Answer: $0.5 \, \text{M}$

Answer: Mass concentration uses mass of solute (g/L), while molarity uses moles of solute (mol/L).

Answer: Molarity is based on the volume of the solution, while molality is based on the mass of the solvent.

Answer: There are 15 g of solute for every 100 g of the solution.

Answer: A solution with a precisely known concentration.

Answer: A volumetric flask.

Answer: 58.44 g.

Answer: $M_1V_1 = M_2V_2$.

Answer: It decreases.

Answer: 2 m.

Answer: 3 N.

Answer: 100 mL.

Answer: 50 g/L.

Answer: Goggles, gloves, and a lab coat.

Answer: 1 M.

Answer: It is based on mass, which does not change with temperature.

Answer: A concentrated solution that is to be diluted to a lower concentration for actual use.

Answer: 5.61 g.

Answer: Yes, it is the molarity multiplied by the number of equivalents.
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