Solutions and Concentration Terms: Molarity and Molality

Fundamental Concepts of Solutions

• A solution is defined as a homogeneous mixture composed of two or more non-reacting components or chemical species.

• The components of a solution are categorized into solutes and solvents:

  • Solute: The component that is dissolved in the solvent, usually present in a smaller quantity.

  • Solvent: The component present in a larger quantity. The solvent determines the physical phase of the entire solution.

• Classification based on the number of components:

  • Binary Solution: Composed of exactly $2$ components (one solute and one solvent).

  • Ternary Solution: Composed of exactly $3$ components.

  • Quaternary Solution: Composed of exactly $4$ components.

  • $n$-component Solution: Composed of $n$ distinct components.

• Phase Determination Principles:

  • If the solute and solvent exist in different physical states, the solvent dictates the phase of the final solution.

  • Example 1 — Sodium Amalgam ($Na(Hg)$):

  • Solute: Mercury ($Hg$), which is a liquid.

  • Solvent: Sodium ($Na$), which is a solid.

  • Phase of Solution: Solid.

  • Example 2 — Zinc Amalgam ($Zn(Hg)$):

  • Solute: Mercury ($Hg$), which is a liquid.

  • Solvent: Zinc ($Zn$), which is a solid.

  • Phase of Solution: Solid.

Classification Based on Solute Saturation

• Unsaturated Solution: A solution in which less than the maximum possible amount of solute is dissolved at a given temperature.

• Saturated Solution: A solution in which the maximum amount of solute is dissolved at a specific temperature under dynamic equilibrium.

• Supersaturated Solution: A solution that holds more than the equilibrium maximum amount of dissolved solute at a specified temperature.

Types of Solutions Based on Physical States

• Gaseous Solutions (Solvent is Gas):

  • Solute Gas, Solvent Gas: Oxygen gas ($O_2$) mixed with Nitrogen gas ($N_2$).

  • Solute Liquid, Solvent Gas: Chloroform ($CHCl_3$) mixed with Nitrogen gas ($N_2$).

  • Solute Solid, Solvent Gas: Camphor mixed with Nitrogen gas ($N_2$).

• Liquid Solutions (Solvent is Liquid):

  • Solute Gas, Solvent Liquid: Oxygen ($O_2$) dissolved in Water ($H_2O$).

  • Solute Liquid, Solvent Liquid: Ethanol ($C_2H_5OH$) dissolved in Water ($H_2O$).

  • Solute Solid, Solvent Liquid: Glucose ($C_6H_{12}O_6$) dissolved in Water ($H_2O$).

• Solid Solutions (Solvent is Solid):

  • Solute Gas, Solvent Solid: Hydrogen gas ($H_2$) in Palladium ($Pd$).

  • Solute Liquid, Solvent Solid: Mercury ($Hg$) in Sodium ($Na$) forming Sodium Amalgam ($Na(Hg)$).

  • Solute Solid, Solvent Solid: Copper ($Cu$) dissolved in Gold ($Au$).

Concentration Terms: Molarity ($M$)

• Definition: Molarity ($M$) is defined as the total number of moles of solute dissolved in one cubic decimeter ($\text{dm}^3$) or liter of solution.

• Mathematical Expression:

$M = \frac{\text{moles of solute}}{\text{volume of solution in }\text{dm}^3}$

$M = \frac{\text{moles of solute}}{\text{volume of solution in }\text{cm}^3} \times 1000$

• Units: $\text{mol composition dm}^{-3}$ or $\text{M}$.

• Temperature Dependence: Molarity is dependent on temperature because the volume of liquid solutions expands or contracts with temperature changes ($V \text{ varies with } T$).

• Conceptual Meaning: A $2 \text{ molar}$ ($2 \text{ M}$) aqueous solution of urea implies that $2 \text{ moles}$ of urea ($NH_2CONH_2$) are dissolved in a solution volume of $1000 \text{ cm}^3$.

• Solved Example 1:

  • Problem: Find the molarity of urea if $30 \text{ g}$ of urea ($NH_2CONH_2$) is dissolved in $100 \text{ cm}^3$ of solution.

  • Molar mass of urea ($NH_2CONH_2$):

$M_{\text{urea}} = 14 + (1 \times 2) + 12 + 16 + 14 + (1 \times 2) = 60 \text{ g mol}^{-1}$

• Calculation:

$\text{moles of urea} = \frac{30 \text{ g}}{60 \text{ g mol}^{-1}} = 0.5 \text{ mol}$

$M = \frac{0.5}{100 \text{ cm}^3} \times 1000 = 5 \text{ M}$

• Solved Example 2:

  • Problem: Find the mass of urea required if the molarity of the solution is $5 \text{ M}$ and the volume of the solution is $250 \text{ cm}^3$.

  • Calculation:

$M = \frac{\text{moles of solute}}{\text{volume of solution in }\text{cm}^3} \times 1000$

$5 = \frac{\frac{x}{60}}{250} \times 1000$

$5 = \frac{x}{60} \times 4$

$5 = \frac{x}{15}$

$x = 5 \times 15 = 75 \text{ g}$

Concentration Terms: Molality ($m$)

• Definition: Molality ($m$) is defined as the number of moles of solute dissolved per kilogram ($\text{kg}$) or $1000 \text{ g}$ of solvent.

• Mathematical Expression:

$m = \frac{\text{moles of solute}}{\text{mass of solvent in }\text{kg}}$

$m = \frac{\text{moles of solute}}{\text{mass of solvent in }\text{g}} \times 1000$

• Units: $\text{mol kg}^{-1}$ or $\text{m}$.

• Temperature Dependence: Molality is independent of temperature because it depends solely on mass, which does not alter with thermal variations.

• Conceptual Meaning: A $2 \text{ molal}$ ($2 \text{ m}$) aqueous solution of urea means $2 \text{ moles}$ of urea are dissolved in $1000 \text{ g}$ of water solvent.

• Solved Example:

  • Problem: Find the mass of $H_2SO_4$ in a $0.25 \text{ m}$ aqueous solution if the mass of the solvent is $250 \text{ g}$.

  • Molar mass of $H_2SO_4$:

$M_{H_2SO_4} = (1 \times 2) + 32 + (16 \times 4) = 98 \text{ g mol}^{-1}$

• Calculation:

$m = \frac{\text{moles of solute}}{\text{mass of solvent in }\text{g}} \times 1000$

$0.25 = \frac{\frac{x}{98}}{250} \times 1000$

$0.25 = \frac{x}{98} \times 4$

$\frac{1}{4} = \frac{4x}{98}$

$16x = 98$

$x = \frac{98}{16} = 6.125 \text{ g}$

Relationship Between Molality ($m$) and Molarity ($M$)

• Conversion Formula:

$m = \frac{1000 \times M}{1000 \times d - M \times M_{\text{solute}}}$

• Where:

  • $m$ = Molality of solution ($\text{mol kg}^{-1}$)

  • $M$ = Molarity of solution ($\text{mol dm}^{-3}$)

  • $d$ = Density of solution ($\text{g cm}^{-3}$)

  • $M_{\text{solute}}$ = Molar mass of solute ($\text{g mol}^{-1}$)

  • Solved Problem:

• Problem: Find out the molality of an $18 \text{ M}$ $H_2SO_4$ aqueous solution if density $d = 1.8 \text{ g cm}^{-3}$.

• Given:

  • $M = 18 \text{ M}$

  • $d = 1.8 \text{ g cm}^{-3}$

  • $M_{\text{solute}} = 98 \text{ g mol}^{-1}$

• Step-by-Step Calculation:

$m = \frac{1000 \times 18}{1000 \times 1.8 - 18 \times 98}$

$m = \frac{18000}{1800 - 1764}$

$m = \frac{18000}{36}$

$m = 500 \text{ m}$

• Application Exercise:

  • Problem: Find out the molarity of a $15 \text{ m}$ urea aqueous solution if density $d = 2 \text{ g cm}^{-3}$.

  • Given:

  • $m = 15 \text{ m}$

  • $d = 2 \text{ g cm}^{-3}$

  • $M_{\text{solute}} (\text{urea}) = 60 \text{ g mol}^{-1}$

  • Step-by-Step Solution:

$15 = \frac{1000 \times M}{1000 \times 2 - M \times 60}$

$15 = \frac{1000 M}{2000 - 60 M}$

$15 \times (2000 - 60 M) = 1000 M$

$30000 - 900 M = 1000 M$

$1900 M = 30000$

$M = \frac{30000}{1900} = \frac{300}{19} \text{ M} \thickapprox 15.79 \text{ M}$