Solved Examples in Solution Chemistry: Molarity, Molality, Density, and Mole Fraction

Calculation of Ionic Concentrations in Mixed Solutions

• Problem Statement: If $20\,mL$ of $0.5\,M$ $Na_{2}SO_{4}$ is mixed with $50\,mL$ of $0.2\,M$ $H_{2}SO_{4}$ and $30\,mL$ of $0.4\,M$ $Al_{2}(SO_{4})_{3}$ solution, calculate the concentrations of $[Na^{+}]$, $[H^{+}]$, $[Al^{3+}]$, and $[SO_{4}^{2-}]$, assuming 100% dissociation.

• Fundamental Formula:     $\text{Molarity} = \frac{\text{moles}}{\text{volume}}$

• Step-by-Step Dissociation and Millimole Calculations:

  • Case (i) $Na_{2}SO_{4}$:
    • Reaction: $Na_{2}SO_{4} \rightarrow 2Na^{+} + SO_{4}^{2-}$
    • Millimoles of $Na_{2}SO_{4} = 20\,mL \times 0.5\,M = 10\,m\,mol$
    • Millimoles of $Na^{+} = 2 \times 10 = 20\,m\,mol$
    • Millimoles of $SO_{4}^{2-} = 10\,m\,mol$
    • Total Volume of final mixture: $20 + 50 + 30 = 100\,mL$
    • $[Na^{+}] = \frac{20\,m\,mol}{100\,mL} = 0.2\,M$
  • Case (ii) $H_{2}SO_{4}$:
    • Reaction: $H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$
    • Millimoles of $H_{2}SO_{4} = 50\,mL \times 0.2\,M = 10\,m\,mol$
    • Millimoles of $H^{+} = 2 \times 10 = 20\,m\,mol$
    • Millimoles of $SO_{4}^{2-} = 10\,m\,mol$
    • $[H^{+}] = \frac{20\,m\,mol}{100\,mL} = 0.2\,M$
  • Case (iii) $Al_{2}(SO_{4})_{3}$:
    • Reaction: $Al_{2}(SO_{4})_{3} \rightarrow 2Al^{3+} + 3SO_{4}^{2-}$
    • Millimoles of $Al_{2}(SO_{4})_{3} = 30\,mL \times 0.4\,M = 12\,m\,mol$
    • Millimoles of $Al^{3+} = 2 \times 12 = 24\,m\,mol$
    • Millimoles of $SO_{4}^{2-} = 3 \times 12 = 36\,m\,mol$
    • $[Al^{3+}] = \frac{24\,m\,mol}{100\,mL} = 0.24\,M$

• Calculation of Total Sulphate Concentration:

  • Total $m\,mol$ of $SO_{4}^{2-} = \text{m.m of } SO_{4}^{2-} \text{ from } (Na_{2}SO_{4} + H_{2}SO_{4} + Al_{2}(SO_{4})_{3})$
  • Total $m\,mol = 10 + 10 + 36 = 56\,m\,mol$
  • $[SO_{4}^{2-}] = \frac{10 + 10 + 36}{100\,mL} = \frac{56}{100} = 0.56\,M$

Determining Molality from Molarity and Density

• Problem Statement: Calculate the molality of an $1.2\,M$ $H_{2}SO_{4}$ solution if its density ($\rho$) is $1.4\,g/mL$.

  • (a) $0.936$
  • (b) $0.562$
  • (c) $0.386$
  • (d) $0.425$

• Solution Procedure:

  • Formula for Molality ($m$) involving Molarity ($M$) and density ($d$):         $m = \frac{M \times 1000}{(1000 \times d) - (M \times M_{solute})}$
  • The molar mass of $H_{2}SO_{4}$ is $98\,g/mol$.
  • Calculation: $m = \frac{1.2 \times 1000}{1000 \times 1.4 - 1.2 \times 98} = \frac{1200}{1400 - 117.6} = \frac{1200}{1282.4} = 0.936$
  • Conclusion: Option (a) is the correct answer.

Calculating Density from Molar Concentration and Mass Percentage

• Problem Statement: A $6.90\,M$ solution of $KOH$ in water has $30\%$ by weight of $KOH$. Calculate the density of the solution (in $g/mL$).

  • (a) $9.345\,g/mL$
  • (b) $2.445\,g/mL$
  • (c) $5.852\,g/mL$
  • (d) $1.288\,g/mL$

• Solution Procedure:

  • Assume a volume of $V = 1\,L$, therefore the moles of solute = $6.9\,mol$.
  • The molar mass of $KOH$ is $56\,g/mol$.
  • Weight of solute = $6.9 \times 56\,gm$.
  • Percentage by mass formula: $\text{\% by mass} = \frac{M_{solute}}{M_{solution}} \times 100$
  • Note that Mass of Solution = $d \times 1000$
  • Equation: $30 = \frac{6.9 \times 56}{d \times 1000} \times 100$
  • Rearranging: $d = \frac{6.9 \times 56 \times 100}{30 \times 1000}$
  • Result: $d = 1.288\,g/mL$
  • Conclusion: Option (d) is the correct answer.

Properties of 3M Sodium Thiosulfate ($Na_{2}S_{2}O_{3}$) Solution

• Given Data:

  • Molarity = $3\,M$
  • Density ($\rho$) = $1.25\,g/mL$
  • Molar mass of solute ($Na_{2}S_{2}O_{3}$) = $158\,g/mol$

• Part (a): Percentage by Weight (w/w):

  • Mass of solution = $1000 \times 1.25 = 1250\,gm$
  • Mass of $Na_{2}S_{2}O_{3} = 3 \times 158 = 474\,gm$
  • $\% (w/w) = \frac{474}{1250} \times 100 = 37.92\%$
  • Result: 37.92 is the correct answer.

• Part (b): Mole Fraction of $Na_{2}S_{2}O_{3}$:

  • Let $X$ be the mole fraction: $X = \frac{n_{solute}}{n_{solute} + n_{solvent}}$
  • Moles of solvent (water) = $\frac{\text{Mass of solution} - \text{Mass of solute}}{\text{Molar mass of water}}$
  • Moles of water = $\frac{1250 - 474}{18} = 43.11\,mol$
  • Mole Fraction ($X_{solute}$) = \frac{3}{3 + 43.11} = 0.065\n * **Result**: 0.065 is the correct answer.\n\n# Molality of Aqueous Urea Solution from Mole Fraction\n\n* **Problem Statement**: Calculate the molality of an aqueous urea solution which has a mole fraction of urea (X_{urea}$) =$0.2.\n * (a) 15.99\n * (b) 13.88\n * (c) 10.99\n * (d) 19.76\n\n* **Solution Procedure**:\n * The sum of mole fractions is 1: X_{u} + X_{w} = 1\n * X_{w} = 1 - 0.2 = 0.8\n * Let total moles = 1\,mol.\n * n_{u} = 0.2\,mol\n * n_{w} = 0.8\,mol\n * Weight of solvent (water): W_{w} = 0.8 \times 18 = 14.4\,gm\n * Molality formula: m = \frac{\text{Number of moles of solute}}{\text{Weight of solvent (kg)}}\n * m = \frac{0.2}{14.4 / 1000} = \frac{0.2 \times 1000}{14.4} = 13.88\,mol/kg\n * **Conclusion**: Option (b) is the correct answer.\n\n# Vapor Pressure and Molar Mass Determination\n\n* **Problem Statement**: 10\,gm$of a solute is dissolved in$80\,gm$of acetone. The vapor pressure of this solution is$271\,mm\,Hg$. If the vapor pressure of pure acetone is$283\,mm\,Hg, calculate the molar mass of the solute.\n * (a) 173\,g/mol\n * (b) 163\,g/mol\n * (c) 143\,g/mol\n * (d) 183\,g/mol\n\n* **Initial Data Parameters**:\n * Weight of solute (w_{B}$) =$10\,g\n * Weight of solvent (acetone, w_{A}$) =$80\,g\n * Vapor pressure of solution (P_{s}$) =$271\,mm\,Hg\n * Vapor pressure of pure solvent (P^{\circ}$) =$283\,mm\,Hg\n * Target: Molar mass of solute (M_{B}$$).