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ISC 2027
All chaptersChemistry · Unit 1

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10 articles35 formulas56 ways the board asks it
CHEExam Practice

Mixed and Comprehensive Questions

These problems combine concentration terms, all four colligative properties, the van't Hoff factor and osmotic-flow direction into multi-step numericals. The key skill is choosing the right formula chain, tracking the factor ii for electrolytes, and keeping units consistent throughout.

Relative lowering of vapour pressure
p∘−psp∘=xsolute=n2n1+n2\dfrac{p^\circ - p_s}{p^\circ} = x_{solute} = \dfrac{n_2}{n_1 + n_2}
p∘p^\circ pure-solvent vapour pressure, psp_s solution vapour pressure, n2n_2 moles solute, n1n_1 moles solvent
Boiling-point elevation and freezing-point depression
ΔTb=iKbmΔTf=iKfm\Delta T_b = i K_b m \qquad \Delta T_f = i K_f m
ii van't Hoff factor, Kb,KfK_b,K_f molal constants, mm molality
Osmotic pressure
π=iCRT\pi = i C R T
CC molarity, R=0.0821R = 0.0821 L atm K−1^{-1} mol−1^{-1}, TT in kelvin
Molar mass from depression / elevation
M2=i K w2×1000ΔT×w1M_2 = \dfrac{i \, K \, w_2 \times 1000}{\Delta T \times w_1}
w2w_2 solute mass (g), w1w_1 solvent mass (g), K=KbK = K_b or KfK_f
van't Hoff factor (dissociation)
i=1+(n−1)αi = 1 + (n-1)\alpha
nn ions per formula unit, α\alpha degree of dissociation
  • Master colligative set: relative lowering p∘−psp∘=xsolute\dfrac{p^\circ - p_s}{p^\circ} = x_{solute}; ΔTb=iKbm\Delta T_b = i K_b m; ΔTf=iKfm\Delta T_f = i K_f m; π=iCRT\pi = iCRT — all depend on number of particles, not their nature.
  • Molality is preferred over molarity in colligative work because it uses mass of solvent and is temperature-independent (volume changes with TT, mass does not).
  • For an electrolyte giving nn ions, i=1+(n−1)αi = 1 + (n-1)\alpha (dissociation) and i=1−(1−1n)αi = 1 - (1 - \tfrac{1}{n})\alpha for association; e.g. Na2SO4→2Na++SO42−Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} has ideal i=3i = 3.
  • Molar mass route from ΔTf\Delta T_f or ΔTb\Delta T_b: M=iK×w2×1000ΔT×w1M = \dfrac{i K \times w_2 \times 1000}{\Delta T \times w_1}, where w2w_2 is solute mass and w1w_1 solvent mass in grams.
  • Osmotic-flow direction: water moves from lower to higher effective particle concentration iCiC, e.g. 0.1 M0.1\,M NaClNaCl (iC=0.2iC = 0.2) is more concentrated in particles than 0.1 M0.1\,M glucose (iC=0.1iC = 0.1), so water flows toward the NaClNaCl.
  • Henry's law link: dissolved gas amount xgas=pgasKHx_{gas} = \dfrac{p_{gas}}{K_H}; convert mole fraction to moles using moles of water in 1 L1\,\text{L} (≈55.5\approx 55.5 mol).
  • Plan multi-part numericals by identifying the measured property first (ΔTf\Delta T_f, ΔTb\Delta T_b, π\pi or psp_s), back-solving for ii or molality, and only then moving to molar mass or degree of dissociation.
  • Percentage dissociation from data: get ii from the observed property, then α=i−1n−1\alpha = \dfrac{i-1}{n-1} and multiply by 100100; e.g. a 0.25 m0.25\,m Na2SO4Na_2SO_4 giving ΔTf=1.08 K\Delta T_f = 1.08\,\text{K} (with Kf=1.86K_f = 1.86) yields i≈2.32i \approx 2.32.
  • A freezing point quoted as −0.93∘C-0.93^\circ\text{C} means ΔTf=0.93 K\Delta T_f = 0.93\,\text{K}; never substitute the negative temperature directly into ΔTf=iKfm\Delta T_f = iK_f m.
  • All four colligative properties scale with the same particle count, so for a given solution the ratios ΔTb:ΔTf:π\Delta T_b : \Delta T_f : \pi are fixed by KbK_b, KfK_f and RTRT — useful for cross-checking an answer.
  • Keep a unit checklist: Kb,KfK_b,K_f in K kg mol−1\text{K kg mol}^{-1}, mm in mol kg−1\text{mol kg}^{-1}, CC in mol L−1\text{mol L}^{-1}, masses converted (solvent to kg for molality, kept in g for the molar-mass formula).
  • Common trap: forgetting ii for ionic solutes inflates the apparent molar mass and underestimates ΔT\Delta T — always check whether the solute is an electrolyte before plugging numbers.
Where the marks go
  • Mixing molality and molarity within one problem — colligative formulae for ΔT\Delta T need molality, but π=iCRT\pi = iCRT needs molarity; converting requires the solution density.
  • Substituting a negative freezing-point temperature into ΔTf=iKfm\Delta T_f = iK_f m instead of its magnitude relative to the pure solvent.
  • Dropping ii for the electrolyte in a multi-step chain, so the back-calculated molar mass comes out abnormally high.
  • Forgetting to convert solution volume from mL to L before using C=n/VC = n/V in the osmotic-pressure step.
  • Using α=i−1\alpha = i - 1 (correct only for a 2-ion salt) instead of the general α=i−1n−1\alpha = \dfrac{i-1}{n-1} for salts giving three or more ions.
How the board asks it
  • Numericalmolar mass from ΔTf=iKfm\Delta T_f = i K_f m
    A solution of 1.25 g1.25\,\text{g} of a non-electrolyte in 20 g20\,\text{g} of water freezes at −1.02 ∘C-1.02\,^\circ\text{C}. Calculate the molar mass of the solute. (Kf=1.86 K kg mol−1K_f = 1.86\,\text{K kg mol}^{-1})
  • Numericalvan't Hoff factor and percentage dissociation from ΔTf\Delta T_f
    0.25 m0.25\,\text{m} aqueous Na2SO4Na_2SO_4 shows a freezing-point depression of 1.08 K1.08\,\text{K}. Taking Kf=1.86 K kg mol−1K_f = 1.86\,\text{K kg mol}^{-1}, calculate the van't Hoff factor ii and the percentage dissociation of the salt.
  • Numericalosmotic pressure π=iCRT\pi = iCRT with mL-to-L conversion
    2.5 g2.5\,\text{g} of K2SO4K_2SO_4 (molar mass 174174) is dissolved to make 250 mL250\,\text{mL} of solution. Assuming complete dissociation, calculate the osmotic pressure at 300 K300\,\text{K}. (R=0.0821 L atm K−1 mol−1R = 0.0821\,\text{L atm K}^{-1}\,\text{mol}^{-1})
  • Numericaladditive colligative effect of a mixture of two non-electrolytes
    0.6 g0.6\,\text{g} of urea (molar mass 6060) and 1.8 g1.8\,\text{g} of glucose (molar mass 180180) are dissolved in 100 g100\,\text{g} of water. If Kb=0.52 K kg mol−1K_b = 0.52\,\text{K kg mol}^{-1}, calculate the elevation in boiling point of the resulting solution.
  • Give reasonseffective particle concentration iCiC and osmotic flow
    Account for the fact that a 0.1 M0.1\,\text{M} NaClNaCl solution exerts nearly twice the osmotic pressure of a 0.1 M0.1\,\text{M} glucose solution at the same temperature.
  • Conversionmolality-to-molarity link via density
    An aqueous solution of glucose (molar mass 180180) is 0.5 m0.5\,\text{m} and has a density of 1.04 g mL−11.04\,\text{g mL}^{-1}. Calculate the molarity of the solution.

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.