Solutions — JEE Mains Chemistry Formulas and Traps

Henry's Law and Solubility of Gases

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}

What the Henry constant depends on

Gas	Temperature	Henry constant in water (kbar)	What it shows
He	293 K	144.97	The largest constant here, so the least soluble gas
$\mathrm{H_2}$	293 K	69.16	About half of helium's constant, so about twice as soluble
$\mathrm{N_2}$	293 K	76.48	Less soluble than oxygen at the same temperature
$\mathrm{N_2}$	303 K	88.84	The constant rises on warming by 10 K, so solubility falls
$\mathrm{O_2}$	293 K	34.86	About 2.2 times as soluble as nitrogen at 293 K
$\mathrm{O_2}$	303 K	46.82	Warmer water holds less oxygen The same gas at two temperatures: the constant is not fixed for a gas.
$\mathrm{CO_2}$	298 K	1.67	A small constant: very soluble, which is why soda water holds so much

A larger constant means a less soluble gas; every gas listed at two temperatures has the larger constant at the higher one.

Common traps

Use the partial pressure, not the total pressure

Henry's law uses the gas's own partial pressure. For a gas that is 20% of air at 5 atm,

p = 1

atm. Putting 5 atm into

p = K_H x

gives a mole fraction five times too large.

Match the pressure unit to KH

If

K_H

is in mmHg, convert the partial pressure from atm to mmHg (

\times 760

) before dividing. Dividing atm by mmHg gives an answer 760 times too small, and that wrong value is usually an option.

KH is not a property of the gas alone

The statement 'KH does not differ for the same gas in different solvents' is false. The constant measures how readily the gas dissolves in that particular liquid, so it changes with the solvent and with temperature.

Warm water holds less gas

Dissolving a gas is exothermic, so raising the temperature drives gas out. The Henry constant rises with temperature and the solubility falls; water near 4 °C holds more oxygen than boiling water.

Raoult's Law for Volatile Liquids

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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}

Positive and negative deviations from Raoult's law

Mixture	Deviation	Reason	Vapour pressure and boiling point
Benzene + toluene	None (ideal)	Similar molecules, similar attractions	On the Raoult line; $\Delta V_{\text{mix}} = 0$
n-Hexane + n-heptane	None (ideal)	Two similar non-polar chains	On the Raoult line; $\Delta H_{\text{mix}} = 0$
Acetone + $\mathrm{CS_2}$	Positive	$\mathrm{CS_2}$ breaks the dipole attraction between acetone molecules	Vapour pressure above the line; boils lower
Ethanol + water	Positive	Ethanol breaks some of water's hydrogen bonds	Minimum-boiling azeotrope, about 95% ethanol by volume
Methanol + $\mathrm{CCl_4}$	Positive	$\mathrm{CCl_4}$ breaks the hydrogen bonds of methanol	Vapour pressure above the line; boils lower
Chloroform + acetone	Negative	The C–H of chloroform hydrogen-bonds to the C=O of acetone	Maximum-boiling azeotrope The standard negative-deviation pair; the answer to 'maximum-boiling azeotrope'.
Acetone + aniline	Negative	The N–H of aniline hydrogen-bonds to the C=O of acetone	Vapour pressure below the line; boils higher
Nitric acid + water	Negative	Strong attraction between the acid and water	Maximum-boiling azeotrope, about 68% nitric acid by mass

A new hydrogen bond between the two liquids means a negative deviation; broken attractions within one liquid mean a positive one.

Common traps

Attach each mole fraction to its own liquid

With 1 mol A and 3 mol B,

x_A = 0.25

multiplies

p^\circ_A

, not

p^\circ_B

. Swapping them gives a pure vapour pressure several times too large, and that value is always among the options.

Higher pure vapour pressure means more volatile

The liquid that escapes more easily has the larger

p^\circ

. A question that asks for the LEAST volatile component wants the one with the smaller

p^\circ

.

The vapour mole fraction needs the total pressure

y_A

is

x_A p^\circ_A

divided by the TOTAL vapour pressure, not by

p^\circ_A

. Stopping at

x_A p^\circ_A

gives a partial pressure, not a mole fraction.

The vapour is richer in the more volatile liquid

The liquid with the larger

p^\circ

always has a bigger share in the vapour than in the liquid. An answer with the vapour poorer in that liquid has a slip in it.

Positive deviation gives the MINIMUM-boiling azeotrope

A positive deviation raises the vapour pressure, so the mixture boils at a lower temperature than either liquid near the azeotrope. Ethanol + water is minimum-boiling; chloroform + acetone is maximum-boiling. Swapping the two is the match-list trap.

A new hydrogen bond means a negative deviation

When mixing makes a hydrogen bond that neither pure liquid had (chloroform with acetone, aniline with acetone), the liquids hold each other more tightly and the vapour pressure falls. When mixing breaks hydrogen bonds (methanol or ethanol diluted by a non-polar liquid), the deviation is positive.

Relative Lowering of Vapour Pressure

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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}

Common traps

Solute's mole fraction, or solvent's?

The relative lowering is the SOLUTE's mole fraction; the solution's vapour pressure divided by

p^\circ

is the SOLVENT's. If the solute's mole fraction is 0.4, the solvent's is 0.6. Read which one the question asks for.

Mass of solution is not mass of solvent

In a w/v solution, 100 mL of solution weighs density times 100 mL, and the solute's mass must be taken out to get the solvent's. Using the solution's full mass overcounts the water.

An electrolyte multiplies the solute's moles

A salt that dissociates gives

i

particles per formula unit. Leaving out

i

for

\mathrm{MgCl_2}

or NaCl gives a lowering that is too small.

Moles of solute come from the elevation, not from a molar mass

When the solute's molar mass is not given, the only route to its moles is

m = \Delta T_b/K_b

times the kilograms of solvent. Look for the molar mass of the SOLVENT instead: it is what turns grams of solvent into moles.

Kilograms for molality, grams for moles of solvent

Molality uses kilograms of solvent; moles of solvent use grams divided by its molar mass. Mixing the two gives an answer off by a factor of 1000.

Elevation of Boiling Point and Depression of Freezing Point

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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}}}

Vapour pressure diagrams and what freezes out

Feature	What happens	Why
Solution's vapour pressure curve	Lies below the pure solvent's curve at every temperature	The non-volatile solute lowers the vapour pressure
Boiling point	The solution reaches 1 atm (760 mmHg) at a higher temperature	Its vapour pressure starts lower, so it must be heated further
Freezing point	The solution meets the solid solvent's curve at a lower temperature	Its lower vapour pressure matches the solid's only at a lower temperature
What freezes out	Pure solid solvent	The solute stays in the liquid 'Only solute molecules solidify' is the planted false statement.
Solution as ice forms	Grows more concentrated and its freezing point keeps falling	Water leaves as ice while the solute stays
Salt on ice at 0 °C	The ice melts and the mixture cools below 0 °C	Brine freezes below 0 °C, a freezing mixture that keeps ice cream frozen

The solution's curve below the solvent's explains both shifts: a higher boiling point and a lower freezing point.

Common traps

The ratio of shifts is the INVERSE ratio of molar masses

For equal masses in equal solvent,

\Delta T \propto 1/M

. If the depressions are in the ratio 1 : 4, the molar masses are in the ratio 4 : 1. Writing 1 : 4 for the masses is the planted option.

Molality is per kilogram of SOLVENT

Divide the solute's moles by the solvent's mass in kg, not by the solution's mass and not by a volume in litres. A solvent given in mL needs its density first.

Answer in the order asked

When a question asks for P and Q 'respectively', the option with the right pair in the wrong order is always present. Match your two values to their letters before choosing.

Ice is pure solvent

On cooling a solution only the solvent freezes. The solute stays behind, so the remaining liquid grows more concentrated. The ice separated is the starting water minus the water still needed to hold the solute at that temperature.

For water, Kf is larger than Kb

Water's

K_f = 1.86

is more than three times its

K_b = 0.52

. A statement that the boiling point of water rises more than its freezing point falls, for the same solution, is false.

Benzene's Kf is larger than water's

Benzene has

K_f = 5.12

K kg/mol against water's 1.86. The same molality freezes benzene almost three times as far below its normal freezing point.

Only the solvent freezes

At the freezing point of a solution, pure solvent crystallises and the solute is left in the liquid. A statement that the solute solidifies, or that both do, is false.

The solution's vapour pressure is lower, not higher

Every colligative effect starts from the lowered vapour pressure. A statement that the solution's vapour pressure is more than the solvent's contradicts Raoult's law.

Osmosis and Osmotic Pressure

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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

Direction of osmosis and reverse osmosis

Situation	What happens	Why
Two solutions across a semipermeable membrane	Solvent flows from the side of lower iC to the side of higher iC	It dilutes the side with more particles
Ions on either side of the membrane	They stay on their own side; no precipitate or colour forms across it	The membrane passes solvent only 'Blue colour forms on both sides' is the planted false option.
Naming the sides	The side with higher iC is hypertonic, the other hypotonic	It has the higher osmotic pressure
Concentrations as osmosis runs	The concentrated side's molarity falls; the dilute side's rises	Water leaves the dilute side and enters the concentrated side
Reverse osmosis	Apply a pressure greater than π on the concentrated side	Pure solvent is pushed back to the dilute side, as in desalination
Membrane for reverse osmosis	Cellophane or parchment paper, not a porous partition	A porous partition lets the solute through as well

The membrane decides what moves (solvent only); the particle count decides which way.

Common traps

Osmotic pressures do not add on mixing

Mixing two solutions of the same concentration gives the same concentration, so the same

\pi

. Adding the two pressures doubles the answer; that doubled value is an option.

Match R to the pressure unit

With

R = 0.083

L bar/(K mol) the pressure must be in bar; with

R = 8.314

it comes out in kPa (litres) or Pa (cubic metres). Put a pressure in Pa into the bar form and the molar mass is off by

10^5

.

Litres of solution, not of solvent

C

is moles per litre of SOLUTION. For a dilute solution 'in 200 mL of water' the two are taken as equal, but the formula itself wants the solution's volume.

Compare iC, not C

0.1 M NaCl and 0.1 M glucose are not isotonic: the salt gives twice the particles. Always multiply each concentration by its ion count before comparing.

Water of crystallisation is not a particle

\mathrm{KCl\cdot MgCl_2\cdot 6H_2O}

gives

\mathrm{K^+}

,

\mathrm{Mg^{2+}}

and three

\mathrm{Cl^-}

: 5 ions. The six water molecules join the solvent and add nothing.

Solvent flows towards the concentrated side

Osmosis moves solvent from the hypotonic to the hypertonic solution, never the other way. A statement that osmosis runs from hypertonic to hypotonic is false.

Reverse osmosis pushes on the concentrated side

The applied pressure must exceed

\pi

and act on the concentrated solution. Pressure on the dilute side only speeds up ordinary osmosis.

Van't Hoff Factor and Abnormal Molar Mass

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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}

Common traps

Ten times the concentration beats twice the ions

0.01 M KCl has

iC = 0.02

; 0.001 M KCl has

0.002

. Two solutions of the same salt are not about equal when one is ten times as concentrated. Compute

i \times C

for each.

Dilution raises a strong electrolyte's i

For NaCl at 0.1, 0.01 and 0.001 M,

i

increases towards 2 as the solution gets more dilute. The reversed order is the planted option.

Divide by n − 1, not by n

\alpha = (i - 1)/(n - 1)

. For

\mathrm{MX_3}

with

i = 1.9

,

\alpha = 0.9/3 = 0.3

, not

0.9/4

. For a two-ion salt it is simply

i - 1

.

Count the ions from the formula

\mathrm{MX_2}

and

\mathrm{A_2B}

give three ions each;

\mathrm{MX_3}

gives four. Using

n = 2

for every salt gives the wrong degree of dissociation.

Observed molar mass is LOWER for dissociation

More particles mean a larger shift and so a smaller apparent molar mass:

M_{\text{obs}} = M_{\text{normal}}/i

. If your observed mass came out larger than the formula mass for a salt, the ratio is upside down.

A dimer halves, it does not vanish

For dimerisation

i = 1 - \alpha/2

, not

1 - \alpha

. With

i = 0.7

,

\alpha = 0.6

; using

1 - \alpha

gives 0.3 and a wrong percentage.

Association raises the apparent molar mass

Fewer particles give a smaller shift, so the molar mass worked out from it is too LARGE: about twice the formula mass for a fully dimerised acid.

Solutions formulas

Henry's Law and Solubility of Gases

Henry's law, p = KH·x

What the Henry constant depends on

Common traps

Raoult's Law for Volatile Liquids

Total vapour pressure of an ideal mixture

Composition of the vapour over an ideal mixture

Positive and negative deviations from Raoult's law

Common traps

Relative Lowering of Vapour Pressure

Relative lowering of vapour pressure equals the solute's mole fraction

Relative lowering of vapour pressure from a boiling-point elevation

Common traps

Elevation of Boiling Point and Depression of Freezing Point

Elevation and depression, ΔT = K·m

Ebullioscopic and cryoscopic constants, Kb and Kf

Vapour pressure diagrams and what freezes out

Common traps

Osmosis and Osmotic Pressure

Osmotic pressure, π = iCRT

Isotonic solutions, equal iC

Direction of osmosis and reverse osmosis

Common traps

Van't Hoff Factor and Abnormal Molar Mass

Ranking solutions by particle concentration, i × m

Van 't Hoff factor for dissociation, i = 1 + (n − 1)α

Van 't Hoff factor for association, i = 1 − (1 − 1/n)α

Common traps

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