Molality to Molarity, Colligative Properties, and Colloids - Study Notes

Molality to Molarity: Conversion and Applications

  • Definition of molality
    • m=n<em>extsolitem</em>extsolventm = \frac{n<em>{ ext{solite}}}{m</em>{ ext{solvent}}} where $m$ is in mol/kg (molality) and $n{ ext{solute}}$ is the number of moles of solute, $m{ ext{solvent}}$ is the mass of the solvent in kilograms.
    • The lecture fixes the solvent mass to 1 kg to make molality directly equal to the number of moles of solute, i.e., $m = n{ ext{solute}}$ when $m{ ext{solvent}} = 1 ext{ kg}$.
  • General steps to convert molality to molarity
    • Determine the number of moles of solute: n<em>extsolute=m</em>extsoluteM<em>extsoluten<em>{ ext{solute}} = \frac{m</em>{ ext{solute}}}{M<em>{ ext{solute}}} where $M{ ext{solute}}$ is the molar mass of the solute.
    • Compute the mass of the solute: m<em>extsolute=n</em>extsoluteimesMextsolute.m<em>{ ext{solute}} = n</em>{ ext{solute}} imes M_{ ext{solute}}.
    • Determine the mass of the solution: m<em>extsolution=m</em>extsolute+mextsolvent.m<em>{ ext{solution}} = m</em>{ ext{solute}} + m_{ ext{solvent}}.
    • Convert to volume using density: V{ ext{solution}} = rac{m{ ext{solution}}}{
      ho{ ext{solution}}} where $ ho{ ext{solution}}$ is in g/mL; convert to liters for molarity.
    • Compute molarity: M=n<em>extsoluteV</em>extsolutionM = \frac{n<em>{ ext{solute}}}{V</em>{ ext{solution}}}
  • Worked example (glucose in water)
    • Given: molality $m = 0.11$ m, solute glucose, MW $M{ ext{glucose}} = 180.2 ext{ g/mol}$, solvent mass $m{ ext{solvent}} = 1.000 ext{ kg}$.
    • Number of moles of glucose: nextglucose=mimes1extkg=0.11extmol.n_{ ext{glucose}} = m imes 1 ext{ kg} = 0.11 ext{ mol}.
    • Mass of glucose: m<em>extglucose=n</em>extglucoseimesMextglucose=0.11imes180.2=19.82extg.m<em>{ ext{glucose}} = n</em>{ ext{glucose}} imes M_{ ext{glucose}} = 0.11 imes 180.2 = 19.82 ext{ g}.
    • Mass of solution: mextsolution=1000extg+19.82extg=1019.82extg.m_{ ext{solution}} = 1000 ext{ g} + 19.82 ext{ g} = 1019.82 ext{ g}.
    • Density (solution) used: $
      ho_{ ext{solution}}
      ightarrow 1.064 ext{ g/mL}$ (approx at these conditions).
    • Volume of solution: Vextsolution=1019.821.064extmL o958.5extmL=0.9585extL.V_{ ext{solution}} = \frac{1019.82}{1.064} ext{ mL} \ o 958.5 ext{ mL} = 0.9585 ext{ L}.
    • Molarity: M=n<em>extglucoseV</em>extsolution=0.110.9585 o0.1148extM.M = \frac{n<em>{ ext{glucose}}}{V</em>{ ext{solution}}} = \frac{0.11}{0.9585} \ o 0.1148 ext{ M}.
  • Key concepts: electrolytes, nonelectrolytes, and the role of particle count
    • Electrolytes ionize in solution and conduct electricity; nonelectrolytes do not ionize (e.g., glucose).
    • Colligative properties depend on the number of dissolved particles, not their identity.
    • Van't Hoff factor $i$ accounts for the effective number of particles formed by dissolution: for nonelectrolytes, $i=1$; for electrolytes, $i>1$ (depending on dissociation).
  • Vapor pressure lowering (Raoult’s law form)
    • Vapor pressure of the solution: P<em>extsolution=P0(1−x</em>extsolute)P<em>{ ext{solution}} = P^{0} \big(1 - x</em>{ ext{solute}}\big) where $P^{0}$ is the vapor pressure of the pure solvent, and $x_{ ext{solute}}$ is the mole fraction of the solute.
    • Alternative algebraic form: P<em>extsolution=P0−x</em>extsoluteP0 extor (P0−P<em>extsolution)=x</em>extsoluteP0.\boxed{P<em>{ ext{solution}} = P^{0} - x</em>{ ext{solute}} P^{0}} \, ext{or}\, \boxed{ \bigl(P^{0} - P<em>{ ext{solution}}\bigr) = x</em>{ ext{solute}} P^{0}}.
    • Mole fraction of solute: x<em>extsolute=n</em>extsoluten<em>extsolute+n</em>extsolvent.x<em>{ ext{solute}} = \frac{n</em>{ ext{solute}}}{n<em>{ ext{solute}} + n</em>{ ext{solvent}}}.
    • This framework yields the change in vapor pressure, $ riangle P = P^{0} - P{ ext{solution}} = x{ ext{solute}} P^{0}$, which is positive (i.e., $P_{ ext{solution}}$ is lower than $P^{0}$).
  • Example: 73.56 g sucrose in 1 L of solution at 100°C
    • $n_{ ext{sucrose}} = rac{73.56}{342.30} approx 0.2149 ext{ mol}$.
    • Density of solution at 100°C: $
      ho{ ext{solution}} = 0.9957 ext{ g/mL}$ → mass of 1 L solution: $m{ ext{solution}} = 995.7 ext{ g}$.
    • Mass of water (solvent): $m_{ ext{water}} = 995.7 - 73.56 = 922.14 ext{ g}$.
    • Moles water: $n_{ ext{water}} = rac{922.14}{18.015} approx 51.17 ext{ mol}$.
    • Mole fraction of sucrose: x_{ ext{sucrose}} = rac{0.2149}{0.2149 + 51.17} approx 0.00418.
    • Vapor pressure of solution: $oxed{P{ ext{solution}} = P^{0}(1 - x{ ext{sucrose}}) = 760 ext{ mmHg}(1 - 0.00418) approx 756.8 ext{ mmHg}}$.
    • Vapor pressure lowering: $ riangle P = x_{ ext{sucrose}} P^{0} approx 3.18 ext{ mmHg}$.
  • Boiling point elevation (BP elevation)
    • Definition: riangleT<em>b=iK</em>bmriangle T<em>b = i K</em>b m where $K_b$ is the ebullioscopic constant and $m$ is molality.
    • For a nonelectrolyte, $i = 1$; for water, $K_b = 0.5652 ext{ °C} ext{ kg/mol}$.
    • Using the example molality $m approx 0.2331 ext{ m}$ (from the glucose example),
    • riangle T_b approx 1 imes 0.5652 imes 0.2331 approx 0.132 ext{ °C}.
    • Thus, $T_{ ext{solution}} approx 100.0 ext{ °C} + 0.132 ext{ °C} approx 100.13 ext{ °C}$.
  • Freezing point depression
    • Definition: riangleT<em>f=iK</em>fmriangle T<em>f = i K</em>f m where $K_f$ is the cryoscopic constant.
    • For water, $K_f = 1.86 ext{ °C kg/mol}$; for nonelectrolyte, $i = 1$.
    • With $m approx 0.2331 ext{ m}$,
    • riangle T_f approx 1.86 imes 0.2331 approx 0.433 ext{ °C}.
    • Freezing point of solution: $T_f( ext{solution}) = 0.0 ext{ °C} - 0.433 ext{ °C} approx -0.433 ext{ °C}.$
  • Ion effects and the Van't Hoff factor
    • For ionic compounds, the solute particles increase due to dissociation; the effective number of particles is given by the Van't Hoff factor $i$.
    • Examples:
    • NaCl → Na⁺ + Cl⁻, so $i approx 2$.
    • CaCl₂ → Ca^{2+} + 2 Cl⁻, so $i approx 3$.
    • When calculating colligative properties for ionic solutes, use $ riangle T = i K m$ (or $P_{ ext{solution}}$ via mole fractions with the effective particle count).
    • Ionic equation reminder: NaCl(s)
      ightarrow Na^{+} + Cl^{-}; CaCl_{2}(s)
      ightarrow Ca^{2+} + 2 Cl^{-}.
  • Quick practical example: 0.05 m CaCl₂
    • $i = 3$ (three particles: Ca^{2+} and 2 Cl⁻).
    • $ riangle Tf = i Kf m = 3 imes 1.86 imes 0.05 approx 0.279^ ext{°C}$.
    • $T_f( ext{solution}) approx 0.0^ ext{°C} - 0.279^ ext{°C} approx -0.279^ ext{°C}$.
  • Molecular weight determination via freezing point depression
    • Given a mass of solute that lowers the freezing point of a solvent by a measured amount, you can back-calculate the molecular weight of the solute.
    • For a nonelectrolyte: compute molality from the depression, then find moles from solvent mass, and finally MW = mass_solute / n.
    • Example: 13.6 g unknown non-electrolyte lowers freezing point of benzene by $ riangle Tf = 0.540^ ext{°C}$; $Kf( ext{benzene}) = 5.12^ ext{°C kg/mol}$; solvent mass $m_{ ext{solvent}} = 2.50 ext{ kg}$.
    • m = rac{ riangle Tf}{Kf} = rac{0.540}{5.12} approx 0.1055 ext{ mol/kg}.
    • n = m imes m_{ ext{solvent}} = 0.1055 imes 2.50 approx 0.2638 ext{ mol}.
    • MW = rac{m_{ ext{sample}}}{n} = rac{13.6}{0.2638} approx 51.6 ext{ g/mol}.$
  • Additional notes on density and units
    • Always track density units to convert mass to volume properly: e.g., $
      ho{ ext{solution}} = 0.9957 ext{ g/mL}$ implies $m{ ext{solution}} =
      ho{ ext{solution}} imes V{ ext{solution}}$ with $V$ in mL.
    • Sign conventions: boiling point elevation yields a positive $ riangle T_b$; freezing point depression yields a negative freezing point for the solution relative to the pure solvent.
    • When applying equations, check that the final temperatures make sense with the physical meaning (e.g., $T{ ext{solution}} > T{ ext{pure}}$ for BP elevation).
  • Quick recap: key formulas to memorize
    • Molality: m=n<em>extsolutem</em>extsolventm = \frac{n<em>{ ext{solute}}}{m</em>{ ext{solvent}}}
    • Molarity: M=n<em>extsoluteV</em>extsolutionM = \frac{n<em>{ ext{solute}}}{V</em>{ ext{solution}}}
    • Number of moles: n<em>extsolute=m</em>extsoluteMextsoluten<em>{ ext{solute}} = \frac{m</em>{ ext{solute}}}{M_{ ext{solute}}}
    • Mass of solution: m<em>extsolution=m</em>extsolute+mextsolventm<em>{ ext{solution}} = m</em>{ ext{solute}} + m_{ ext{solvent}}
    • Volume from density: V{ ext{solution}} = rac{m{ ext{solution}}}{
      ho_{ ext{solution}}}
    • Vapor pressure lowering: P<em>extsolution=P0(1−x</em>extsolute)extwithx<em>extsolute=n</em>extsoluten<em>extsolute+n</em>extsolventP<em>{ ext{solution}} = P^{0}\big(1 - x</em>{ ext{solute}}\big) ext{ with } x<em>{ ext{solute}} = \frac{n</em>{ ext{solute}}}{n<em>{ ext{solute}} + n</em>{ ext{solvent}}}
    • Boiling point elevation: riangleT<em>b=iK</em>bmriangle T<em>b = i K</em>b m
    • Freezing point depression: riangleT<em>f=iK</em>fmriangle T<em>f = i K</em>f m
    • For nonelectrolytes: $i = 1$; for electrolytes, adjust $i$ via the number of particles (Van't Hoff factor).
    • Molecular weight via freezing point: $MW = rac{m{ ext{solute}}}{n}$ where $n = m imes m{ ext{solvent}}$ and $m = rac{ riangle Tf}{i Kf}$.
  • Final practical note
    • The same set of concepts applies across problems, just keep track of whether the solute is an electrolyte (affecting $i$) and carefully convert between moles, masses, volumes, and densities.
    • The instructor emphasizes practice with signs, unit consistency, and using given constants (e.g., $Kb$, $Kf$, and density) as provided in problems.