Solutions — JEE Mains Chemistry

Formulas (1)

• Henry's law, p = KH·x · Henry's law
  $$p = K_H\,x \qquad n_{\text{gas}} \approx x \times 55.56 \text{ mol per litre of water}$$

Reference tables (1)

Watch out for (4)

• Use the partial pressure, not the total pressure
  → Henry's law, p = KH·x
• Match the pressure unit to KH
  → Henry's law, p = KH·x
• KH is not a property of the gas alone
  → What the Henry constant depends on
• Warm water holds less gas
  → What the Henry constant depends on

Formulas (2)

• Total vapour pressure of an ideal mixture · Raoult's law for two volatile liquids
  $$P = x_A p^\circ_A + x_B p^\circ_B = p^\circ_B + \left(p^\circ_A - p^\circ_B\right)x_A$$
• Composition of the vapour over an ideal mixture · Vapour mole fraction
  $$y_A = \frac{x_A p^\circ_A}{P} \qquad \frac{1}{P} = \frac{y_A}{p^\circ_A} + \frac{y_B}{p^\circ_B}$$

Reference tables (1)

Watch out for (6)

• Attach each mole fraction to its own liquid
  → Total vapour pressure of an ideal mixture
• Higher pure vapour pressure means more volatile
  → Total vapour pressure of an ideal mixture
• The vapour mole fraction needs the total pressure
  → Composition of the vapour over an ideal mixture
• The vapour is richer in the more volatile liquid
  → Composition of the vapour over an ideal mixture
• Positive deviation gives the MINIMUM-boiling azeotrope
  → Positive and negative deviations from Raoult's law
• A new hydrogen bond means a negative deviation
  → Positive and negative deviations from Raoult's law

Formulas (2)

• Relative lowering of vapour pressure equals the solute's mole fraction · Raoult's law for a non-volatile solute
  $$\frac{p^\circ - p}{p^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$$
• Relative lowering of vapour pressure from a boiling-point elevation · Linking the two colligative effects
  $$\frac{p^\circ - p}{p^\circ} \approx \frac{m\,M_1}{1000} = \frac{\Delta T_b\,M_1}{1000\,K_b}$$

Watch out for (5)

• Solute's mole fraction, or solvent's?
  → Relative lowering of vapour pressure equals the solute's mole fraction
• Mass of solution is not mass of solvent
  → Relative lowering of vapour pressure equals the solute's mole fraction
• An electrolyte multiplies the solute's moles
  → Relative lowering of vapour pressure equals the solute's mole fraction
• Moles of solute come from the elevation, not from a molar mass
  → Relative lowering of vapour pressure from a boiling-point elevation
• Kilograms for molality, grams for moles of solvent
  → Relative lowering of vapour pressure from a boiling-point elevation

Formulas (2)

• Elevation and depression, ΔT = K·m · Colligative temperature shifts
  $$\Delta T_b = K_b\,m \qquad \Delta T_f = K_f\,m \qquad M_2 = \frac{1000\,K\,w_2}{\Delta T\,W_1}$$
• Ebullioscopic and cryoscopic constants, Kb and Kf · The solvent constants
  $$K_b = \frac{R\,T_b^{2}\,M_1}{1000\,\Delta H_{\text{vap}}} \qquad K_f = \frac{R\,T_f^{2}\,M_1}{1000\,\Delta H_{\text{fus}}} = \frac{M_1 R\,T_f}{1000\,\Delta S_{\text{fus}}}$$

Reference tables (1)

Watch out for (8)

• The ratio of shifts is the INVERSE ratio of molar masses
  → Elevation and depression, ΔT = K·m
• Molality is per kilogram of SOLVENT
  → Elevation and depression, ΔT = K·m
• Answer in the order asked
  → Elevation and depression, ΔT = K·m
• Ice is pure solvent
  → Elevation and depression, ΔT = K·m
• For water, Kf is larger than Kb
  → Ebullioscopic and cryoscopic constants, Kb and Kf
• Benzene's Kf is larger than water's
  → Ebullioscopic and cryoscopic constants, Kb and Kf
• Only the solvent freezes
  → Vapour pressure diagrams and what freezes out
• The solution's vapour pressure is lower, not higher
  → Vapour pressure diagrams and what freezes out

Formulas (2)

• Osmotic pressure, π = iCRT · Van 't Hoff equation for osmotic pressure
  $$\pi = iCRT \qquad M = \frac{wRT}{\pi V} \qquad \pi = h\rho g$$
• Isotonic solutions, equal iC · Isotonic condition
  $$i_1 C_1 = i_2 C_2$$

Reference tables (1)

Watch out for (7)

• Osmotic pressures do not add on mixing
  → Osmotic pressure, π = iCRT
• Match R to the pressure unit
  → Osmotic pressure, π = iCRT
• Litres of solution, not of solvent
  → Osmotic pressure, π = iCRT
• Compare iC, not C
  → Isotonic solutions, equal iC
• Water of crystallisation is not a particle
  → Isotonic solutions, equal iC
• Solvent flows towards the concentrated side
  → Direction of osmosis and reverse osmosis
• Reverse osmosis pushes on the concentrated side
  → Direction of osmosis and reverse osmosis

Formulas (3)

• Ranking solutions by particle concentration, i × m · Colligative effects scale with particle concentration
  $$\Delta T_b = i K_b m \qquad \Delta T_f = i K_f m \qquad \pi = iCRT$$
• Van 't Hoff factor for dissociation, i = 1 + (n − 1)α · Degree of dissociation
  $$i = 1 + (n - 1)\alpha \qquad \alpha = \frac{i - 1}{n - 1} \qquad K_a = \frac{C\alpha^2}{1 - \alpha}$$
• Van 't Hoff factor for association, i = 1 − (1 − 1/n)α · Degree of association
  $$i = 1 - \left(1 - \frac{1}{n}\right)\alpha \qquad \text{dimer: } i = 1 - \frac{\alpha}{2}$$

Watch out for (7)

• Ten times the concentration beats twice the ions
  → Ranking solutions by particle concentration, i × m
• Dilution raises a strong electrolyte's i
  → Ranking solutions by particle concentration, i × m
• Divide by n − 1, not by n
  → Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
• Count the ions from the formula
  → Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
• Observed molar mass is LOWER for dissociation
  → Van 't Hoff factor for dissociation, i = 1 + (n − 1)α
• A dimer halves, it does not vanish
  → Van 't Hoff factor for association, i = 1 − (1 − 1/n)α
• Association raises the apparent molar mass
  → Van 't Hoff factor for association, i = 1 − (1 − 1/n)α