Solutions

Fundamentals of Solutions and Classification

• Definition of Solution: A solution is defined as a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.

• Binary Solution: A solution consisting of exactly two components is called a binary solution.

  • Solute: The component present in a relatively smaller amount in the solution.
  • Solvent: The component present in a relatively larger amount in the solution. The physical state of the solvent determines the physical state of the overall solution.
  • Examples of Binary Solutions:
  • $\text{NaCl} + \text{H}_2\text{O}$
  • $\text{Glucose} + \text{H}_2\text{O}$
  • $\text{C}_2\text{H}_5\text{OH} + \text{H}_2\text{O}$
  • $\text{Urea} + \text{H}_2\text{O}$

• Classification Based on Solute Concentration:

  • Dilute Solutions: Solutions that contain a very small amount of solute relative to the volume of the solvent.
  • Example: A solution containing $0.5$\,mol of $\text{NaCl}$ in $1$\,L of $\text{H}_2\text{O}$.
  • Concentrated Solutions: Solutions that contain a significant amount of solute relative to the volume of the solvent.
  • Example: A solution containing $0.8$\,mol of $\text{NaCl}$ in $1$\,L of $\text{H}_2\text{O}$.

Concentration Terms and Expressing Composition

• Definition of Concentration: Concentration is the amount of solute present in a given amount (mass or volume) of solution or solvent.

• Classification of Concentration Terms by Temperature Dependence:

  • Temperature-Dependent Concentration Terms: Terms involving liquid volume. Since volume varies with temperature, these concentration terms change when temperature changes.
  • Percent Mass by Volume ($\%\,\text{w/v}$)
  • Percent Volume by Volume ($\%\,\text{v/v}$)
  • Molarity ($M$)
  • Normality ($N$)
  • Temperature-Independent Concentration Terms: Terms involving mass or moles only, without any volume terms. These remain constant regardless of temperature changes.
  • Percent Mass by Mass ($\%\,\text{w/w}$)
  • Mole Fraction ($X$)
  • Molality ($m$)

• Percent Mass by Mass ($\%\,\text{w/w}$):

  • Definition: Mass of solute in grams present per $100$\,g of solution.
  • Formula:     $\%\,\text{w/w} = \frac{\text{Mass of solute (g)}}{\text{Mass of solution (g)}} \times 100 = \frac{\text{Mass of solute (g)}}{\text{Mass of solute (g)} + \text{Mass of solvent (g)}} \times 100$
  • Example: $20$\,g of $\text{NaOH}$ is dissolved in $80$\,g of $\text{H}_2\text{O}$. Determine the $\%\,\text{w/w}$.
  • Mass of solute = $20$\,g
  • Mass of solution = $20$\,g + $80$\,g = $100$\,g
  • $\%\,\text{w/w} = \frac{20}{100} \times 100 = 20\%\,\text{w/w}$

• Percent Mass by Volume ($\%\,\text{w/v}$):

  • Definition: Mass of solute in grams present per $100$\,mL of solution.
  • Formula:     $\%\,\text{w/v} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100$
  • Interconversion Example 1: A solution is $30\%\,\text{w/w}$ having a density of $1.20$\,g/mL. Determine its $\%\,\text{w/v}$.
  • $30\%\,\text{w/w}$ means $30$\,g of solute is present in $100$\,g of solution.
  • Density of solution $d = \frac{m}{V} \rightarrow V_{\text{sol}} = \frac{100}{1.2}$\,mL.
  • $\%\,\text{w/v} = \frac{30}{\frac{100}{1.2}} \times 100 = 30 \times 1.2 = 36\%\,\text{w/v}$.
  • Interconversion Example 2: An aqueous solution of glucose is $5.4\%\,\text{w/v}$. If the density of the solution is $1.8$\,g/mL, determine its $\%\,\text{w/w}$.
  • Formula relation: $\%\,\text{w/w} = \frac{\%\,\text{w/v}}{d_{\text{sol}}}$
  • $\%\,\text{w/w} = \frac{5.4}{1.8} = 3\%\,\text{w/w}$.

• Percent Volume by Volume ($\%\,\text{v/v}$):

  • Definition: Volume of solute in mL present per $100$\,mL of solution.
  • Formula:     $\%\,\text{v/v} = \frac{\text{Volume of solute (mL)}}{\text{Volume of solution (mL)}} \times 100$
  • Example: Find the volume of solvent present in $500$\,mL of a $30\%\,\text{v/v}$ solution.
  • $30 = \frac{\text{Volume of solute}}{500} \times 100 \rightarrow \text{Volume of solute} = 150$\,mL.
  • Volume of solvent = $500$\,mL - $150$\,mL = $350$\,mL.

• Molarity ($M$):

  • Definition: The total number of moles of solute present in $1$\,L ($1000$\,mL) of solution.
  • Formulas:     $M = \frac{\text{Number of moles of solute } (n_B)}{\text{Volume of solution in liters } (V_{\text{L}})}$$M = \frac{W_B \times 1000}{M_B \times V_{\text{mL}}}$     where $W_B$ is mass of solute in grams, $M_B$ is molar mass of solute in g/mol, and $V_{\text{mL}}$ is solution volume in mL.
  • Units: $\text{mol\,L}^{-1}$ or $\text{M}$.
  • Terminology:
  • 1 Molar (1 M): Solution containing $1$\,mol of solute in $1$\,L solution.
  • Semimolar (0.5 M or 1/2 M): Solution containing $0.5$\,mol of solute in $1$\,L solution.
  • Decimolar (0.1 M or 1/10 M): Solution containing $0.1$\,mol of solute in $1$\,L solution.
  • Numerical Problems:
  • Problem 1: $0.2$\,mol of solute is present in $2$\,L of solution. Find molarity.       $M = \frac{0.2}{2} = 0.1$\,M.
  • Problem 2: $0.5$\,mol of solute is present in $400$\,mL of solution. Find molarity.       $M = \frac{0.5 \times 1000}{400} = 1.25$\,M.
  • Problem 3: $4$\,g of $\text{NaOH}$ ($M_B = 23 + 16 + 1 = 40$\,g/mol) is dissolved in $100$\,cm$^3$ of $\text{H}_2\text{O}$. Find molarity (given $1$\,cm$^3 = 1$\,mL).       $M = \frac{4 \times 1000}{40 \times 100} = 1$\,M.
  • Problem 4: How many grams of $\text{H}_2\text{SO}_4$ ($M_B = 98$\,g/mol) are required to prepare $500$\,mL of $0.2$\,M $\text{H}_2\text{SO}_4$ solution?       $0.2 = \frac{W_B \times 1000}{98 \times 500} \rightarrow W_B = 0.2 \times 49 = 9.8$\,g.
  • Problem 5: An aqueous solution of $\text{NaOH}$ has a strength of $4$\,g/L. Determine its molarity.       $\text{Molarity} = \frac{\text{Strength (g/L)}}{M_B (\text{g/mol})} = \frac{4}{40} = 0.1$\,mol/L.
  • Problem 6: An aqueous solution of glucose ($M_B = 180$\,g/mol) is $5.4\%\,\text{w/v}$. Determine its molarity.       $\text{Molarity} = \frac{\%\,\text{w/v} \times 10}{M_B} = \frac{5.4 \times 10}{180} = 0.3$\,M.
  • Problem 7: An aqueous solution of $\text{KOH}$ ($M_B = 56$\,g/mol) is $28\%\,\text{w/w}$ having a density of $1.25$\,g/mL. Determine its molarity.       $\text{Molarity} = \frac{\%\,\text{w/w} \times d_{\text{sol}} \times 10}{M_B} = \frac{28 \times 1.25 \times 10}{56} = 6.25$\,M.
  • Problem 8: Find the molarity of pure $\text{H}_2\text{O}$ (density = $1$\,g/mL).       For $1$\,L ($1000$\,mL) of $\text{H}_2\text{O}$, mass = $1000$\,g.       $n_{\text{H}_2\text{O}} = \frac{1000}{18} = 55.55$\,mol.       $\text{Molarity} = \frac{55.55\,\text{mol}}{1\,\text{L}} = 55.55$\,M.
  • Problem 9: Find molarity of $720$\,g of $\text{H}_2\text{O}$.       Molarity is an intensive property (independent of total mass or volume). Thus, molarity remains $55.55$\,M.

• Dilution of Solutions:

  • Definition: Adding extra solvent to an existing solution is called dilution.
  • During dilution, the total amount of solute remains unchanged ($n_1 = n_2$).
  • Dilution Formula:     $M_1 V_1 = M_2 V_2$     where $V_2 = V_1 + V_{\text{solvent added}}$.
  • Problem 1: $200$\,mL of $0.1$\,M $\text{H}_2\text{SO}_4$ solution is diluted with $300$\,mL of $\text{H}_2\text{O}$. Find molarity of resulting solution.     $V_2 = 200 + 300 = 500$\,mL.     $0.1 \times 200 = M_2 \times 500 \rightarrow M_2 = \frac{20}{500} = 0.04$\,M.
  • Problem 2: $400$\,mL of $0.5$\,M $\text{NaOH}$ solution is diluted with $\text{H}_2\text{O}$. If the molarity of the resulting solution is $0.05$\,M, determine the volume of $\text{H}_2\text{O}$ added.     $V_2 = \frac{M_1 V_1}{M_2} = \frac{0.5 \times 400}{0.05} = 4000$\,mL.     $\text{Volume of } \text{H}_2\text{O} \text{ added} = V_2 - V_1 = 4000 - 400 = 3600$\,mL or $3.6$\,L.

• Molarity of Non-Reacting Mixtures:

  • When solutions containing the same solute are mixed:     $n_3 = n_1 + n_2 \rightarrow M_3 V_3 = M_1 V_1 + M_2 V_2$$M_3 = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$
  • Problem 1: $200$\,mL of $0.1$\,M $\text{HCl}$ solution is mixed with $300$\,mL of $0.1$\,M $\text{HCl}$ solution. Find molarity of resulting solution.     $M_3 = \frac{(0.1 \times 200) + (0.1 \times 300)}{200 + 300} = \frac{20 + 30}{500} = 0.1$\,M.
  • Problem 2: $400$\,mL of $0.2$\,M $\text{NaOH}$ solution is mixed with $600$\,mL of $0.05$\,M $\text{NaOH}$ solution. Find molarity of resulting solution.     $M_3 = \frac{(0.2 \times 400) + (0.05 \times 600)}{400 + 600} = \frac{80 + 30}{1000} = 0.11$\,M.

• Molarity of Reacting Mixtures:

  • Neutralization reaction between an acid and a base (e.g., $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$).
  • The resulting molarity is determined by the moles of Excess Reactant (ER) left over after Limiting Reactant (LR) is completely consumed:     $M_3 V_3 = |M_1 V_1 - M_2 V_2| \rightarrow M_3 = \frac{|M_1 V_1 - M_2 V_2|}{V_1 + V_2}$
  • Problem: $600$\,mL of $0.1$\,M $\text{HCl}$ solution is mixed with $400$\,mL of $0.2$\,M $\text{NaOH}$ solution. Find molarity of resulting solution.     $n_{\text{HCl}} = 0.1 \times 600 = 60$\,mmol, $n_{\text{NaOH}} = 0.2 \times 400 = 80$\,mmol.     $\text{NaOH}$ is the excess reactant. Moles of $\text{NaOH}$ left = $80 - 60 = 20$\,mmol.     $M_3 = \frac{80 - 60}{600 + 400} = \frac{20}{1000} = 0.08$\,M.

• Mole Fraction ($X$):

  • Definition: The ratio of the number of moles of one component to the total number of moles of all components present in the solution.
  • Formulas:     $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$$X_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solute}} + n_{\text{solvent}}}$$X_{\text{solute}} + X_{\text{solvent}} = 1$
  • Properties: It is a dimensionless, unitless quantity.
  • Problem 1: $30$\,g of urea ($M_B = 60$\,g/mol) is present in $45$\,g of $\text{H}_2\text{O}$. Find mole fraction of urea and $\text{H}_2\text{O}$.
  • $n_{\text{urea}} = \frac{30}{60} = 0.5$\,mol
  • $n_{\text{H}_2\text{O}} = \frac{45}{18} = 2.5$\,mol
  • Total moles = $0.5 + 2.5 = 3.0$\,mol
  • $X_{\text{urea}} = \frac{0.5}{3.0} = \frac{1}{6}$
  • $X_{\text{H}_2\text{O}} = 1 - \frac{1}{6} = \frac{5}{6}$
  • Problem 2: An aqueous solution of $\text{KOH}$ is $28\%\,\text{w/w}$. Find mole fraction of $\text{KOH}$ ($M_B = 56$\,g/mol).
  • $28$\,g $\text{KOH}$ in $72$\,g $\text{H}_2\text{O}$.
  • $n_{\text{KOH}} = \frac{28}{56} = 0.5$\,mol
  • $n_{\text{H}_2\text{O}} = \frac{72}{18} = 4$\,mol
  • Total moles = $4.5$\,mol
  • $X_{\text{KOH}} = \frac{0.5}{4.5} = \frac{1}{9}$, $X_{\text{H}_2\text{O}} = \frac{8}{9}$.
  • Problem 3: Equal masses ($64$\,g each) of $\text{He}$, $\text{CH}_4$, $\text{O}_2$, and $\text{SO}_2$ are present in a container. Determine their mole fractions.
  • Molar masses: $\text{He} = 4$, $\text{CH}_4 = 16$, $\text{O}_2 = 32$, $\text{SO}_2 = 64$\,g/mol.
  • $n_{\text{He}} = \frac{64}{4} = 16$\,mol
  • $n_{\text{CH}_4} = \frac{64}{16} = 4$\,mol
  • $n_{\text{O}_2} = \frac{64}{32} = 2$\,mol
  • $n_{\text{SO}_2} = \frac{64}{64} = 1$\,mol
  • Total moles = $16 + 4 + 2 + 1 = 23$\,mol
  • $X_{\text{He}} = \frac{16}{23}$, $X_{\text{CH}_4} = \frac{4}{23}$, $X_{\text{O}_2} = \frac{2}{23}$, $X_{\text{SO}_2} = \frac{1}{23}$.

• Molality ($m$):

  • Definition: Number of moles of solute present per $1$\,kg ($1000$\,g) of solvent.
  • Formula:     $m = \frac{\text{Moles of solute } (n_B)}{\text{Mass of solvent in kg } (W_A)} = \frac{W_B \times 1000}{M_B \times W_A(\text{g})}$
  • Units: $\text{mol\,kg}^{-1}$ or molal ($\text{m}$).
  • Temperature independent term.
  • Mole fraction of solute in 1 molal aqueous solution:
  • $1$\,molal means $1$\,mol solute in $1000$\,g $\text{H}_2\text{O}$ ($n_{\text{H}_2\text{O}} = \frac{1000}{18} = 55.55$\,mol).
  • $X_{\text{solute}} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx 0.0177$
  • Relation between Molality and Molarity:     $m = \frac{1000 M}{1000 d - M M_B}$     where $d$ is density of solution in g/mL.
  • Problem 1: An aqueous solution of $\text{CH}_3\text{COOH}$ is $30\%\,\text{w/w}$. Determine its molality ($M_B = 60$\,g/mol).
  • $30$\,g solute in $70$\,g $\text{H}_2\text{O}$.
  • $m = \frac{30 \times 1000}{60 \times 70} = \frac{50}{7}$\text{m}.
  • Problem 2: An aqueous solution of $\text{H}_2\text{SO}_4$ is $18$\,M with density $1.8$\,g/mL. Determine its molality ($M_B = 98$\,g/mol).
  • Mass of $1000$\,mL solution = $1000 \times 1.8 = 1800$\,g.
  • Mass of solute = $18 \times 98 = 1764$\,g.
  • Mass of solvent = $1800 - 1764 = 36$\,g.
  • $m = \frac{18 \times 1000}{36} = 500$\,m.
  • Problem 3: An aqueous solution of urea ($M_B = 60$\,g/mol) is $12$\,M with density $1.2$\,g/mL. Determine its molality.
  • $m = \frac{1000 \times 12}{1000(1.2) - 12(60)} = \frac{12000}{1200 - 720} = \frac{12000}{480} = 25$\,m.
  • Problem 4: Mole fraction of solute in aqueous solution is $0.2$. Find molality.
  • $X_B = 0.2 \rightarrow X_A = 0.8$
  • $m = \frac{X_B \times 1000}{X_A \times M_A} = \frac{0.2 \times 1000}{0.8 \times 18} = \frac{1000}{72} = 13.88$\,m.

• Normality ($N$):

  • Definition: Number of gram-equivalents of solute present in $1$\,L of solution. Temperature dependent quantity.
  • Formulas:     $N = \frac{\text{Number of g-equivalents of solute}}{V_{\text{L}}}$$\text{Number of g-equivalents} = \frac{\text{Given mass}}{\text{Equivalent mass}}$$\text{Equivalent mass} = \frac{\text{Molar mass}}{\text{n-factor}}$
  • n-factor calculation rules:
  • Elements: Valency.
  • Acids: Basicity (number of replaceable $\text{H}^+$ ions).
  • Bases: Acidity (number of $\text{OH}^-$ ions donated by a base).
  • Salts: Total positive charge on cations or total negative charge on anions.
  • Ions: Absolute charge magnitude on the ion.
  • Relation between Molarity and Normality:     $N = M \times \text{n-factor}$
  • Problem 1: Normality of $0.2$\,M $\text{Na}_2\text{CO}_3$ solution?
  • $\text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-} \rightarrow \text{n-factor} = 2$.
  • $N = 0.2 \times 2 = 0.4$\,N.
  • Problem 2: Find molarity of $0.2$\,N $\text{H}_3\text{PO}_4$ solution (assuming complete dissociation, n-factor = 3).
  • $M = \frac{N}{\text{n-factor}} = \frac{0.2}{3} = 0.067$\,M.
  • Problem 3: Determine n-factor of $\text{H}_3\text{PO}_4$ in the reaction: $\text{H}_3\text{PO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{HPO}_4 + 2\text{H}_2\text{O}$.
  • Since $2$\,$\text{H}^+$ ions are replaced, \text{n-factor} = 2$.\n\n- **Parts Per Million (PPM)**:\n - **Formula**:\n    \text{PPM} = \frac{\text{Number of parts of solute}}{\text{Number of parts of solution}} \times 10^6\n - *Problem*: 1$\,mg of solute is present in$1\,kg of solution. Find PPM.\n - \text{PPM} = \frac{10^{-3}\,\text{g}}{10^3\,\text{g}} \times 10^6 = 1\,ppm.\n\n- **Volume Strength of Hydrogen Peroxide (\text{H}2 ext{O}_2)**:\n - **Definition**: Volume of oxygen gas produced at STP when 1$\,L of$\text{H}_2\text{O}_2 solution is completely decomposed upon heating.\n - Chemical reaction: \text{H}_2\text{O}_2 (aq) \xrightarrow{\Delta} \text{H}_2\text{O} (l) + \frac{1}{2}\text{O}_2 (g)\n - **Formulas**:\n - At pressure P = 1$\,atm:$\text{Volume Strength} = M \times 11.2\n - At pressure P = 1$\,bar:$\text{Volume Strength} = M \times 11.35\n - In terms of normality: \text{Volume Strength} = N \times 5.6\n - *Example*: An aqueous solution of \text{H}_2\text{O}_2$is$30\text{V}$. This means$30$\,L of oxygen gas will be produced at STP when$1$\,L of this$\text{H}_2\text{O}_2 solution is heated.\n\n# Vapour Pressure and Phase Equilibrium\n\n- **Definition of Vapour Pressure**: The pressure exerted by vapours present above a liquid surface in a closed container at a condition of dynamic equilibrium is called vapour pressure.\n- **Dynamic Equilibrium**: At equilibrium, \text{Rate of vaporisation} = \text{Rate of condensation}.\n - Example: \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_2\text{O} (g) \rightarrow K_p = P{\text{H}2\text{O}} = (VP){ ext{sol}}\n\n- **Types of Substances Based on Vapour Pressure**:\n - **Volatile Substance**: A substance having measurable vapour pressure (VP eq 0).\n - *Examples*: \text{H}2\text{O}$, n-hexane, n-heptane,$\text{CCl}_4$,$\text{CHCl}_3$,$\text{CH}_3\text{COCH}_3$,$\text{CH}_3\text{CH}_2\text{OH}.\n - **Non-Volatile Substance**: A substance having zero or negligible vapour pressure (VP = 0).\n - *Examples*: \text{NaCl} (s), glucose, urea, sucrose.\n\n- **Factors Affecting Vapour Pressure**:\n - **Nature of Liquid**: Vapour pressure is inversely proportional to intermolecular forces of attraction.\n - \text{CH}_3\text{CH}_2\text{OH}$(ethyl alcohol, BP =$78.39^\circ$\,C) has higher$VP$than$\text{H}_2\text{O}$(BP =$100^\circ\,C) at the same temperature because ethyl alcohol has weaker intermolecular hydrogen bonding.\n - n-Hexane vs n-Heptane: n-Hexane has higher VP than n-heptane due to smaller molecular size and weaker van der Waals forces.\n - Polyhydroxy alcohols: VP$follows the order: Ethyl alcohol > Ethylene glycol > Glycerol, because glycerol contains three$-\text{OH} groups leading to maximum intermolecular hydrogen bonding.\n - **Temperature**: Vapour pressure of a pure liquid increases with temperature as described by the Clausius-Clapeyron equation:\n    \frac{d\ln(VP)}{dT} = \frac{\Delta H{\text{vap}}}{R T^2}\n\n# Raoult's Law and Vapour Pressure of Solutions\n\n- **Statement of Raoult's Law**: The partial vapour pressure of any volatile component in a solution is directly proportional to its mole fraction in that solution.\n\n- **Case I: When a Non-Volatile Solute (B) is added to a Volatile Liquid Solvent (A)**:\n - Solute B is solid, non-volatile (P_B^\circ = 0$,$P_B = 0).\n - Solvent A is liquid, volatile (P_A^\circ eq 0).\n - Total solution vapour pressure P_T = P_A = P_A^\circ X_A = P_A^\circ (1 - X_B).\n - **Lowering in Vapour Pressure**: \Delta P = P_A^\circ - P_A\n - **Relative Lowering in Vapour Pressure (RLVP)**:\n    \text{RLVP} = \frac{P_A^\circ - P_A}{P_A^\circ} = X_B\n - *Problem*: 30$\,g of urea ($M_B = 60$\,g/mol) is mixed with$45$\,g of$\text{H}2\text{O}$($P{\text{H}_2\text{O}}^\circ = 300\,mmHg). Determine:\n 1. \%\,\text{w/w} = \frac{30}{75} \times 100 = 40\%\n 2. Mole fraction of urea = \frac{0.5}{0.5 + 2.5} = \frac{0.5}{3.0} = \frac{1}{6}\n 3. Vapour pressure of solution P_A = 300 \times \left(1 - \frac{1}{6}\right) = 300 \times \frac{5}{6} = 250\,mmHg\n 4. Lowering in VP = 300 - 250 = 50\,mmHg\n 5. \text{RLVP} = \frac{50}{300} = \frac{1}{6}\n 6. Molality = \frac{0.5 \times 1000}{45} = 11.11\,m\n\n- **Case II: Liquid-Liquid Solutions (Two Volatile Liquids A and B)**:\n - Solute B is volatile (P_B^\circ eq 0$), Solvent A is volatile ($P_A^\circ eq 0).\n - Partial pressure of A: P_A = P_A^\circ X_A\n - Partial pressure of B: P_B = P_B^\circ X_B\n - Total Vapour Pressure (P_T):\n    P_T = P_A + P_B = P_A^\circ X_A + P_B^\circ X_B\n - Alternate linear form: P_T = P_A^\circ + (P_B^\circ - P_A^\circ) X_B\n - *Problem 1*: Solution of 2$\,mol A and$3$\,mol B.$P_A^\circ = 200$\,mmHg,$P_B^\circ = 300$\,mmHg. Find$P_T$.     $X_A = \frac{2}{5}$, $X_B = \frac{3}{5}$.     $P_T = 200\left(\frac{2}{5}\right) + 300\left(\frac{3}{5}\right) = 80 + 180 = 260$\,mmHg.
  • Problem 2: Two volatile liquids A and B with $P_A^\circ = 400$\,torr, $P_B^\circ = 500$\,torr.
  • Case a: Equal moles of A and B mixed ($X_A = X_B = 0.5$):       $P_T = \frac{P_A^\circ + P_B^\circ}{2} = \frac{400 + 500}{2} = 450$\,torr.
  • Case b: Equal masses of A and B mixed ($M_A = 20$, $M_B = 40$):       Let mass = $W$\,g. $n_A = \frac{W}{20}$, n_B = \frac{W}{40} \rightarrow n_A = 2 n_B$.\n      X_A = \frac{2}{3}$,$X_B = \frac{1}{3}$.       $P_T = 400\left(\frac{2}{3}\right) + 500\left(\frac{1}{3}\right) = \frac{800 + 500}{3} = \frac{1300}{3} = 433.33$\,torr.
  • Problem 3: Two volatile liquids A and B having $P_A^\circ = 108$\,torr, $P_B^\circ = 36$\,torr form a solution with total pressure $P_T = 90$\,torr. Find mole fractions.     90 = 108 + (36 - 108) X_B \rightarrow -18 = -72 X_B \rightarrow X_B = \frac{18}{72} = 0.25$, X_A = 0.75$.

• Mole Fraction in Vapour Phase ($Y_A, Y_B$):

  • By Dalton's Law of Partial Pressures: $P_A = Y_A P_T$ and $P_B = Y_B P_T$.
  • Formulas:     $Y_A = \frac{P_A}{P_T} = \frac{P_A^\circ X_A}{P_A^\circ X_A + P_B^\circ X_B}$$Y_B = \frac{P_B}{P_T} = \frac{P_B^\circ X_B}{P_A^\circ X_A + P_B^\circ X_B}$
  • Problem: Benzene (B) and Toluene (A) solution. Mole fraction of Toluene in liquid solution $X_A = 0.4 \rightarrow X_B = 0.6$. Given $P_{\text{toluene}}^\circ = 170$\,torr, $P_{\text{benzene}}^\circ = 240$\,torr. Determine mole fraction of benzene in vapour phase ($Y_B$).     $P_T = (240 \times 0.6) + (170 \times 0.4) = 144 + 68 = 212$\,torr.     (If taking Toluene as component A and Benzene as component B with $240 \times 0.4 + 170 \times 0.6 = 96 + 102 = 198$\,torr):     $Y_A = \frac{96}{198} = \frac{16}{33}$, $Y_B = \frac{102}{198} = \frac{17}{33}$.

Ideal vs Non-Ideal Solutions

• Molecular Interaction Framework:

  • $A-A$ interaction: Forces between solvent molecules.
  • $B-B$ interaction: Forces between solute molecules.
  • $A-B$ interaction: Forces between solvent and solute molecules in solution.

• Ideal Solutions:

  • Solutions that obey Raoult's law at all concentrations and temperatures.
  • Characteristics:
  • $A-B$ interaction strength is equal to $A-A$ and $B-B$ interactions.
  • $\Delta V_{\text{mix}} = 0$ (volume of solution equals sum of component volumes, $V_1 + V_2 = V_3$).
  • $\Delta H_{\text{mix}} = 0$ (no enthalpy change during mixing).
  • $\Delta S_{\text{mix}} > 0$ (entropy increases upon mixing).
  • $\Delta G_{\text{mix}} < 0$ (mixing is spontaneous).
  • $P_{\text{obs}} = P_{\text{calculated}} = P_A^\circ X_A + P_B^\circ X_B$
  • Examples:
  • n-Hexane + n-Heptane
  • Benzene + Toluene
  • $\text{CCl}_4 + \text{SiCl}_4$
  • $\text{C}_2\text{H}_5\text{Cl} + \text{C}_2\text{H}_5\text{Br}$
  • $\text{C}_2\text{H}_4\text{Cl}_2 + \text{C}_2\text{H}_4\text{Br}_2$

• Non-Ideal Solutions:

  • Solutions that do not obey Raoult's law over the entire range of concentrations.
  • Characteristics:
  • $A-B$ interaction strength is not equal to $A-A$ and $B-B$ interactions.
  • $\Delta V_{\text{mix}} \neq 0$
  • $\Delta H_{\text{mix}} \neq 0$
  • $\Delta S_{\text{mix}} > 0$
  • $\Delta G_{\text{mix}} < 0$

• Types of Non-Ideal Solutions:

  • Solutions Showing Positive Deviation from Raoult's Law:
  • $A-B$ interactions are weaker than $A-A$ and $B-B$ interactions.
  • Solute and solvent molecules escape more easily into the vapour phase.
  • $P_{\text{obs}} > P_{\text{calculated}}$
  • $P_A > P_A^\circ X_A$ and $P_B > P_B^\circ X_B$
  • $\Delta V_{\text{sol}} > 0$ (expansion occurs on mixing)
  • $\Delta H_{\text{sol}} > 0$ (endothermic dissolution process)
  • Examples:
    • $\text{H}_2\text{O} + \text{Benzene}$
    • $\text{H}_2\text{O} + \text{Toluene}$
    • $\text{CCl}_4 + \text{Benzene}$
    • $\text{H}_2\text{O} + \text{Ethanol} (\text{C}_2\text{H}_5\text{OH})$
    • $\text{H}_2\text{O} + \text{n-hexane}$
    • $\text{Benzene} + \text{Methanol}$
  • Solutions Showing Negative Deviation from Raoult's Law:
  • $A-B$ interactions are stronger than $A-A$ and $B-B$ interactions.
  • Solute and solvent molecules are held more tightly, reducing escaping tendency.
  • $P_{\text{obs}} < P_{\text{calculated}}$
  • $P_A < P_A^\circ X_A$ and $P_B < P_B^\circ X_B$
  • $\Delta V_{\text{sol}} < 0$ (contraction occurs on mixing)
  • $\Delta H_{\text{sol}} < 0$ (exothermic dissolution process)
  • Examples:
    • $\text{CHCl}_3 + \text{CH}_3\text{COCH}_3$ (chloroform + acetone, due to strong hydrogen bonding between them)
    • $\text{HNO}_3 + \text{H}_2\text{O}$
    • $\text{HCl} + \text{H}_2\text{O}$
    • $\text{H}_2\text{SO}_4 + \text{H}_2\text{O}$

Azeotropic Mixtures (Azeotropes)

• Definition: Binary liquid mixtures that boil at a constant temperature and distill over completely without any change in liquid or vapour composition.

• Key Characteristic: The components of an azeotropic mixture cannot be separated by fractional distillation.

• Rule: Ideal solutions do not form azeotropes. Only non-ideal solutions form azeotropic mixtures.

• Types of Azeotropes:

  • Minimum Boiling Azeotrope:
  • Formed by solutions showing positive deviation from Raoult's law.
  • The boiling point of the azeotrope is lower than the boiling points of both individual pure components.
  • Example: $\text{H}_2\text{O}$ (BP = $100^\circ$\,C) + $\text{C}_2\text{H}_5\text{OH}$ (BP = $78.37^\circ$\,C) forms a minimum boiling azeotrope at $96\%$ ethanol + $4\%$ water with BP = $78.15^\circ$\,C.
  • Maximum Boiling Azeotrope:
  • Formed by solutions showing negative deviation from Raoult's law.
  • The boiling point of the azeotrope is higher than the boiling points of both individual pure components.
  • Example: $\text{H}_2\text{O}$ (BP = $$