JEE Corner — Equivalent Concept, Vapour Density, Yield, Purity and Concentration Interconversions

Beyond the Textbook: The JEE Toolkit

JEE Main uses several terms the Class 11 syllabus never defines. This section builds them: the equivalent, vapour density, yield and purity, POAC, ppm and ppb, and the relations between molarity, molality, mole fraction and mass per cent.

Why 1 mol of $\mathrm{H_2SO_4}$ is worth 2 mol of $\mathrm{NaOH}$

1 mol of $\mathrm{H_2SO_4}$ neutralises 2 mol of $\mathrm{NaOH}$; 1 mol of $\mathrm{HCl}$ neutralises 1 mol. Moles do not react one-to-one; reactive units (protons, hydroxide ions, electrons) do. Equivalents count those units.

Key Point (Definition): The n-factor of a species is the number of reactive units it supplies or accepts per formula unit in a given reaction. The equivalent mass is $$E = \frac{\text{Molar mass } (M)}{n\text{-factor}}$$ and the number of gram equivalents in a sample is $$\text{Equivalents} = \frac{\text{mass in g}}{E} = \text{moles} \times n\text{-factor}$$

Everything below is about finding the right n-factor.

Acids: n-factor = basicity

Acid	Basicity ($n$)	$M$ (g mol$^{-1}$)	$E$ (g eq$^{-1}$)
$\mathrm{HCl}$, $\mathrm{HNO_3}$, $\mathrm{CH_3COOH}$	1	36.5, 63, 60	36.5, 63, 60
$\mathrm{H_2SO_4}$	2	98	49
$\mathrm{H_2C_2O_4 \cdot 2H_2O}$ (oxalic acid)	2	126	63
$\mathrm{H_3PO_4}$	3	98	32.7
$\mathrm{H_3PO_3}$ (phosphorous acid)	2	82	41
$\mathrm{H_3PO_2}$ (hypophosphorous acid)	1	66	66

Basicity counts only hydrogens bonded to oxygen. $\mathrm{H_3PO_3}$ has two $\mathrm{P-OH}$ and one $\mathrm{P-H}$, so it is dibasic; $\mathrm{H_3PO_2}$ has one $\mathrm{P-OH}$ and two $\mathrm{P-H}$, so it is monobasic. Acetic acid has basicity 1 for the same reason.

The n-factor depends on the reaction. In $\mathrm{H_2SO_4 + NaOH \rightarrow NaHSO_4 + H_2O}$ the acid is half-neutralised, $n = 1$ and $E = 98$.

Bases: n-factor = acidity

$\mathrm{NaOH}$, $\mathrm{KOH}$, $\mathrm{NH_4OH}$: $n = 1$. $\mathrm{Ca(OH)_2}$, $\mathrm{Ba(OH)_2}$: $n = 2$, so $E$ of $\mathrm{Ca(OH)_2} = 74/2 = 37$. $\mathrm{Al(OH)_3}$: $n = 3$, $E = 78/3 = 26$.

Salts (non-redox): n-factor = total positive charge

Salt	Cations	$n$	$E$
$\mathrm{NaCl}$	$1 \times (+1)$	1	58.5
$\mathrm{Na_2CO_3}$	$2 \times (+1)$	2	$106/2 = 53$
$\mathrm{CaCO_3}$	$1 \times (+2)$	2	$100/2 = 50$
$\mathrm{Al_2(SO_4)_3}$	$2 \times (+3)$	6	$342/6 = 57$
$\mathrm{Ca_3(PO_4)_2}$	$3 \times (+2)$	6	$310/6 = 51.7$

Redox species: n-factor = total change in oxidation number per formula unit

Key Point (The KMnO$_4$ rule): The same oxidant has different n-factors in different media.

Medium	Half-reaction	$n$	$E$ ($M = 158$)
Acidic	$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$	5	$158/5 = 31.6$
Neutral / weakly basic	$\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}$	3	$158/3 = 52.7$
Strongly basic	$\mathrm{MnO_4^- + e^- \rightarrow MnO_4^{2-}}$	1	$158/1 = 158$

Colours: purple to colourless (acidic), brown $\mathrm{MnO_2}$ (neutral), green manganate (basic).

Potassium dichromate in acid: $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$. Two Cr atoms each drop $+6 \rightarrow +3$, so $n = 2 \times 3 = 6$ and $E = 294/6 = 49$; $n$ is per formula unit.

Common reductants: $\mathrm{FeSO_4}$ ($\mathrm{Fe^{2+} \rightarrow Fe^{3+}}$, $n = 1$, $E = 152$); Mohr's salt $\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$ ($n = 1$, $E = 392$); oxalic acid as reductant ($\mathrm{C_2O_4^{2-} \rightarrow 2CO_2}$, C goes $+3 \rightarrow +4$ twice, $n = 2$, $E = 63$); $\mathrm{FeC_2O_4}$ (Fe gives 1, oxalate 2, $n = 3$, $E = 144/3 = 48$); $\mathrm{H_2O_2}$ as oxidant or reductant ($n = 2$, $E = 17$).

Key Point: For a species that changes in two places (like $\mathrm{FeC_2O_4}$), add the changes. For disproportionation, treat the oxidised and reduced halves separately; write both half-reactions first.

Normality and the Law of Equivalents

Normality counts equivalents per litre, and equivalents react one-to-one, so titration arithmetic becomes simple.

Key Point (Definition): The normality $N$ of a solution is the number of gram equivalents of solute per litre of solution: $$N = \frac{\text{equivalents of solute}}{\text{volume in L}} = \frac{\text{mass in g}}{E \times V(\text{L})}$$ Since equivalents $=$ moles $\times$ $n$-factor, $$N = n\text{-factor} \times M$$

1 M $\mathrm{H_2SO_4}$ is 2 N. 0.1 M $\mathrm{KMnO_4}$ is 0.5 N in acidic, 0.3 N in neutral and 0.1 N in strongly basic medium — same bottle, three normalities, which is why a normality is only meaningful alongside the reaction.

Normality is never smaller than molarity

Since $n \geq 1$, $N \geq M$ always, with equality only for monobasic acids, monoacidic bases and one-electron redox species.

The law of equivalents

Key Point (Law of Equivalents): In any reaction, reactants combine and products form in equal numbers of equivalents: $$\text{Equivalents of A} = \text{Equivalents of B} = \text{Equivalents of product}$$

No balanced equation is needed; the n-factors carry the stoichiometry. For two solutions reacting completely, equivalents $= N \times V$, so

$$N_1 V_1 = N_2 V_2$$ The volumes may be in any unit, provided both sides use the same one.

Form of reagent	Equivalents
Solution	$N \times V(\text{L})$, or $n \times M \times V(\text{L})$
Pure substance of mass $w$	$w / E$, or $(w/M) \times n$
Gas at STP (1 atm, 22.4 L)	$\dfrac{V(\text{L})}{22.4} \times n$

For hydrogen liberated by a metal ($\mathrm{H^+ + e^- \rightarrow \frac{1}{2}H_2}$), one mole of $\mathrm{H_2}$ is 2 equivalents, so 1 equivalent of any metal liberates 11.2 L of $\mathrm{H_2}$ at STP (1 atm), or 11.35 L at 1 bar. Use 22.7 L only when the question says 1 bar.

Back-titration and mixtures

Back-titration: excess reagent A is added to the sample and the leftover A is titrated with B. $$\text{Equivalents of sample} = \text{Equivalents of A added} - \text{Equivalents of B used}$$

Two acids (or two reductants) titrated together: $$N_{\text{base}} V_{\text{base}} = N_1 V_1 + N_2 V_2$$

[JEE Main] The titrant's n-factor is fixed by the medium: "acidified $\mathrm{KMnO_4}$" means $n = 5$. When only a volume ratio is asked, $\dfrac{V_1}{V_2} = \dfrac{N_2}{N_1}$; equimolar $\mathrm{H_2SO_4}$ and $\mathrm{NaOH}$ react in a 1 : 2 volume ratio.

Vapour Density and Relative Density of Gases

Early chemists compared gases by weighing equal volumes. That comparison is the vapour density, usually a disguised molar mass in a question.

Key Point (Definition): The vapour density (VD) of a gas is the ratio of the mass of a certain volume of the gas to the mass of the same volume of hydrogen at the same temperature and pressure: $$\text{VD} = \frac{\text{mass of } V \text{ litres of gas}}{\text{mass of } V \text{ litres of } \mathrm{H_2}} \quad (\text{same } T, P)$$

VD has no units.

Deriving VD $= M/2$

1. By Avogadro's law, equal volumes of gases at the same $T$ and $P$ contain equal numbers of molecules, $N$.
2. The mass of $N$ molecules of the gas is $N \times m_{\text{gas}}$, and of $N$ molecules of hydrogen is $N \times m_{\mathrm{H_2}}$, so $$\text{VD} = \frac{N\, m_{\text{gas}}}{N\, m_{\mathrm{H_2}}} = \frac{m_{\text{gas}}}{m_{\mathrm{H_2}}} = \frac{M_{\text{gas}}}{M_{\mathrm{H_2}}}$$
3. Hydrogen is diatomic, $M_{\mathrm{H_2}} = 2$ g mol$^{-1}$ (2.016 more precisely), so $$\boxed{\text{VD} = \frac{M}{2} \qquad \text{or} \qquad M = 2 \times \text{VD}}$$

Gas	$M$	VD	Gas	$M$	VD
$\mathrm{H_2}$	2	1	$\mathrm{N_2}$, CO	28	14
$\mathrm{CH_4}$	16	8	$\mathrm{O_2}$	32	16
$\mathrm{NH_3}$	17	8.5	$\mathrm{CO_2}$, $\mathrm{N_2O}$	44	22
$\mathrm{H_2O}$(g)	18	9	$\mathrm{SO_2}$	64	32

Relative density with respect to other gases

"$x$ times as dense as air" means $M = 29x$ (air $\approx 29$); "$y$ times as dense as oxygen" means $M = 32y$. In general

$$\frac{d_{\text{gas}}}{d_{\text{ref}}} = \frac{M_{\text{gas}}}{M_{\text{ref}}} \quad (\text{same } T, P)$$

because $d = \dfrac{PM}{RT}$ and $P/RT$ is common. The factor of 2 belongs to hydrogen only.

Vapour density of a mixture

VD $= M_{\text{avg}}/2$, with $M_{\text{avg}}$ the mole-fraction-weighted average molar mass. Equal moles of $\mathrm{H_2}$ and $\mathrm{O_2}$ give $M_{\text{avg}} = 17$, VD $= 8.5$; for equal masses, convert to moles first.

Degree of dissociation from vapour density

For $\mathrm{A \rightleftharpoons nB}$, molecules multiply, so density falls. Let $D$ be the theoretical VD of A, $d$ the observed VD, $\alpha$ the degree of dissociation. From 1 mol of A: $1 - \alpha$ mol of A and $n\alpha$ mol of B, total $1 + (n-1)\alpha$ mol. Density is inversely proportional to moles, so

$$\frac{D}{d} = 1 + (n - 1)\alpha \qquad \Rightarrow \qquad \boxed{\alpha = \frac{D - d}{(n - 1)\,d}}$$

For $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$ and $\mathrm{N_2O_4 \rightleftharpoons 2NO_2}$, $n = 2$ and $\alpha = (D - d)/d$.

[JEE Main] "Vapour density 32" means $M = 64$; convert at once. "Density at STP in g/L" means $M/22.4$ (or $M/22.7$ at 1 bar).

Per Cent Yield and Per Cent Purity

Per cent yield

Key Point (Definition): The theoretical yield is the maximum product (mass or moles) the balanced equation predicts from the limiting reagent. The actual yield is what the experiment delivers. $$\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Use grams or moles for both, consistently. Yield falls below 100% through side reactions, equilibrium, losses and unreacted material; a yield above 100% is a red flag for an impure or wet product, never a bonus.

Working backwards

For 50 g of product at 80% yield,

$$\text{Theoretical yield needed} = \frac{\text{required actual yield}}{\%\ \text{yield}/100} = \frac{50}{0.80} = 62.5\ \text{g}$$

Divide by the yield fraction going backwards, multiply going forwards.

Sequential reactions: yields multiply

For $\mathrm{A \rightarrow B \rightarrow C \rightarrow D}$ with step yields $y_1, y_2, y_3$ (as fractions),

$$\text{Overall yield} = y_1 \times y_2 \times y_3$$

Steps at 80%, 75% and 50% give $0.80 \times 0.75 \times 0.50 = 0.30$. Yields never add or average.

Key Point: If the stoichiometry changes between steps (1 mol of B gives 2 mol of C), multiply the mole ratio in too: moles of D $=$ moles of A $\times$ (mole ratios) $\times$ (yield fractions).

Per cent purity

Key Point (Definition): The per cent purity of a sample is the mass percentage of pure, reactive substance in it: $$\%\ \text{purity} = \frac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100$$

Assume the impurity is inert and no product is lost. Work back from the product to the mass of pure reactant.

Example: 10 g of impure $\mathrm{CaCO_3}$ gives 3.52 g of $\mathrm{CO_2}$. $\mathrm{CaCO_3 \rightarrow CaO + CO_2}$, so moles of $\mathrm{CaCO_3} =$ moles of $\mathrm{CO_2} = 3.52/44 = 0.080$; pure $\mathrm{CaCO_3} = 0.080 \times 100 = 8.0$ g; purity $= 80\%$.

Yield and purity together

$$\text{Actual product} = (\text{sample mass}) \times \frac{\text{purity}}{100} \times (\text{stoichiometric factor}) \times \frac{\text{yield}}{100}$$

Purity applies to the reactant, yield to the product. Never add the percentages.

Concept	Formula	Applied to
% yield	actual / theoretical $\times$ 100	product, at the end
% purity	pure / impure $\times$ 100	reactant, at the start
Overall yield	product of step yields	whole route
Reactant needed for a target	target $\div$ (yield fraction)	working backwards

[JEE Main] A reagent 20% in excess is taken at 1.2 times the amount the balanced equation demands, not 1.2 times the other reagent's mass.

POAC: The Principle of Atom Conservation

POAC lets you skip balancing.

Key Point (Principle of Atom Conservation): Atoms are neither created nor destroyed in a reaction. For every element, the total moles of its atoms on the reactant side equals the total on the product side, whether or not the equation is balanced.

The recipe

1. Write the reaction with only the species that matter, no coefficients.
2. Pick an element that appears in exactly one reactant of interest and one product of interest.
3. Write (moles of that element on the left) $=$ (moles on the right), using $$\text{moles of atoms of X in a species} = (\text{moles of species}) \times (\text{number of X atoms per formula unit})$$
4. Solve for the unknown moles, then convert to mass or volume.

Worked case

Problem. How many grams of $\mathrm{O_2}$ are obtained by completely decomposing 24.5 g of $\mathrm{KClO_3}$?

$$\mathrm{KClO_3 \rightarrow KCl + O_2} \qquad (\text{unbalanced})$$

Moles of $\mathrm{KClO_3} = \dfrac{24.5}{122.5} = 0.200$ mol. All O goes into $\mathrm{O_2}$, so POAC on oxygen: $$3 \times n_{\mathrm{KClO_3}} = 2 \times n_{\mathrm{O_2}} \quad \Rightarrow \quad n_{\mathrm{O_2}} = \frac{3 \times 0.200}{2} = 0.300\ \text{mol}$$

Mass of $\mathrm{O_2} = 0.300 \times 32 = 9.6$ g, the same as from the balanced $\mathrm{2KClO_3 \rightarrow 2KCl + 3O_2}$.

For KCl, use potassium: $n_{\mathrm{KCl}} = n_{\mathrm{KClO_3}} = 0.200$ mol $= 14.9$ g.

Where POAC is most useful

• Conversions with no reagents given. "What mass of $\mathrm{Fe_2O_3}$ from 240 g of $\mathrm{FeS_2}$?" Iron is conserved. $\mathrm{FeS_2}$ has one Fe per formula unit, so $n_{\mathrm{Fe}} = \frac{240}{120} = 2$ mol, and since $\mathrm{Fe_2O_3}$ holds two, $n_{\mathrm{Fe_2O_3}} = 1$ mol $= 160$ g.
• Large equations where balancing takes longer than the problem.
• Solution reactions where one element matters (mass of $\mathrm{BaSO_4}$ from a sulphate mixture: POAC on sulphur).
• Combustion analysis: moles of C $=$ moles of $\mathrm{CO_2}$; moles of H $= 2 \times$ moles of $\mathrm{H_2O}$.

Two conditions

1. The element must not be split between two products (or two reactants) unless the split is known. In a reaction giving both CO and $\mathrm{CO_2}$, carbon needs the ratio.
2. The reaction must go to completion. POAC counts atoms, not yield; apply the yield fraction afterwards.

Key Point: POAC and the law of equivalents both skip balancing. Use POAC when the question is about an element's mass in a product; use equivalents for a titration or a redox reaction between two reagents.

ppm, ppb and the Concentration Interconversion Web

Any concentration unit can be converted to any other, with density as the connector. Two more units cover trace amounts.

Parts per million and parts per billion

Key Point (Definition): $$\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6}, \qquad \text{ppb} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9}$$

Both are mass-by-mass unless stated otherwise. 1 ppm $= 1$ mg per kg, and dilute aqueous solutions have $d \approx 1$ g mL$^{-1}$, so

$$1\ \text{ppm} \approx 1\ \mathrm{mg\ L^{-1}} \approx 1\ \mathrm{mg\ kg^{-1}}, \qquad 1\ \text{ppb} \approx 1\ \mu\mathrm{g\ L^{-1}}$$

1% $= 10^4$ ppm; 1 ppm $= 1000$ ppb. Water hardness in ppm is the mass of $\mathrm{CaCO_3}$ equivalent per $10^6$ g of water; fluoride in drinking water is about 1 ppm.

The web

Symbols: solute molar mass $M_2$, solvent molar mass $M_1$ (18 for water), molarity $M$, molality $m$, density $d$ in g mL$^{-1}$, mass per cent $w$, solute mole fraction $x_2$.

Route 1: mass per cent to molarity

Basis 100 g of solution: $w/M_2$ mol of solute in $100/d$ mL. $$M = \frac{w/M_2}{100/(1000d)} = \boxed{\frac{10\, d\, w}{M_2}}$$ Check: 98% $\mathrm{H_2SO_4}$, $d = 1.84$, gives $M = 10 \times 1.84 \times 98/98 = 18.4$ M.

Route 2: mass per cent to molality

Same basis; solvent $= (100 - w)$ g. $$m = \frac{w/M_2}{(100 - w)/1000} = \boxed{\frac{1000\, w}{M_2 (100 - w)}}$$ No density needed.

Route 3: molarity to molality

Basis 1 L of solution: $M$ mol of solute ($M M_2$ g) in $1000d$ g total, so solvent $= (1000d - M M_2)$ g. $$m = \frac{M}{(1000d - M M_2)/1000} = \boxed{\frac{1000\, M}{1000\, d - M\, M_2}}$$ Inverse, from 1 kg of solvent (total mass $1000 + m M_2$ g, volume $(1000 + m M_2)/d$ mL): $$M = \frac{1000\, d\, m}{1000 + m\, M_2}$$ For dilute aqueous solutions $d \approx 1$ and $M M_2 \ll 1000$, so $m \approx M$.

Route 4: molality to mole fraction

Basis 1 kg of solvent: $m$ mol solute, $1000/M_1$ mol solvent. $$x_2 = \frac{m}{m + 1000/M_1} = \boxed{\frac{m\, M_1}{m\, M_1 + 1000}}$$ For water, $x_2 = \dfrac{18m}{18m + 1000}$; a 1 molal solution has $x_2 = 18/1018 = 0.0177$. Inverse: $m = \dfrac{1000\, x_2}{x_1 M_1}$, $x_1 = 1 - x_2$.

Route 5: molarity to mole fraction

From 1 L of solution, $n_2 = M$ and $n_1 = (1000d - M M_2)/M_1$: $$x_2 = \frac{M\, M_1}{M\, M_1 + 1000\, d - M\, M_2}$$

The master table

From $\downarrow$ to $\rightarrow$	Molarity $M$	Molality $m$	Mole fraction $x_2$
Mass % $w$	$\dfrac{10\,d\,w}{M_2}$	$\dfrac{1000\,w}{M_2(100-w)}$	$\dfrac{w/M_2}{w/M_2 + (100-w)/M_1}$
Molarity $M$	—	$\dfrac{1000M}{1000d - MM_2}$	$\dfrac{MM_1}{MM_1 + 1000d - MM_2}$
Molality $m$	$\dfrac{1000\,d\,m}{1000 + mM_2}$	—	$\dfrac{mM_1}{mM_1 + 1000}$

Also: strength in g L$^{-1}$ $= M \times M_2$, and $N = n \times M$, so strength $= N \times E$.

Key Point: Do not memorise the table. Pick a basis (100 g of solution, 1 L of solution, or 1 kg of solvent), write the grams and moles of solute and solvent, and read off the unit asked.

[JEE Main] Density is the only bridge between volume-based units (molarity, normality, strength) and mass-based ones (mass %, molality, mole fraction, ppm). If no density is given and the question crosses that boundary, either the solution is meant to be dilute ($d = 1$ g mL$^{-1}$) or the conversion cannot be done at all.

Mixing, Dilution, Average Molar Mass, and Common Traps

Dilution

Solvent changes the volume, not the moles of solute: $$M_1 V_1 = M_2 V_2 \qquad (\text{and identically } N_1 V_1 = N_2 V_2)$$ Water added is $V_2 - V_1$, not $V_2$.

Mixing two solutions of the same solute

Moles add and (assuming additive volumes) volumes add: $$M_{\text{mix}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$$ The result always lies between $M_1$ and $M_2$ — a quick sanity check.

Mixing an acid with a base

Work in equivalents: $$N_{\text{resultant}} = \frac{|N_{\text{acid}} V_{\text{acid}} - N_{\text{base}} V_{\text{base}}|}{V_{\text{acid}} + V_{\text{base}}}$$ Acidic if the acid term is larger. For $\mathrm{H^+}$ or $\mathrm{OH^-}$, $N = M$.

Common ion from different solutes

For $\mathrm{[Cl^-]}$ after mixing $\mathrm{NaCl}$ and $\mathrm{CaCl_2}$, count the ion: $\mathrm{[Cl^-]} = \dfrac{M_1 V_1 \times 1 + M_2 V_2 \times 2}{V_1 + V_2}$.

Average molar mass of a gas mixture

Key Point (Definition): For gases with mole fractions $x_i$ and molar masses $M_i$, $$M_{\text{avg}} = \sum x_i M_i = \frac{\text{total mass}}{\text{total moles}} = \frac{n_1 M_1 + n_2 M_2 + \cdots}{n_1 + n_2 + \cdots}$$ At the same $T$ and $P$, mole fraction $=$ volume fraction. VD of the mixture $= M_{\text{avg}}/2$.

Air: $0.79 \times 28 + 0.21 \times 32 = 22.1 + 6.7 = 28.8 \approx 29$.

Equal moles of $\mathrm{H_2}$ and $\mathrm{O_2}$: $M_{\text{avg}} = (2 + 32)/2 = 17$. Equal masses $w$: moles $w/2$ and $w/32$, so $$M_{\text{avg}} = \frac{2w}{w/2 + w/32} = \frac{2}{17/32} = \frac{64}{17} = 3.76$$ For equal masses of two gases, $M_{\text{avg}} = \dfrac{2 M_1 M_2}{M_1 + M_2}$, the harmonic mean.

Common traps

#	Trap	The fix
1	$n = 5$ for $\mathrm{KMnO_4}$ in every medium	acidic 5, neutral 3, strongly basic 1
2	All H atoms counted as acidic ($\mathrm{H_3PO_3}$, $\mathrm{CH_3COOH}$)	only $\mathrm{O-H}$ hydrogens; $\mathrm{H_3PO_3}$ dibasic, $\mathrm{H_3PO_2}$ monobasic
3	$N < M$	$N = n \times M$; never divide
4	"VD $= 32$ g/L"	VD is unitless; $M = 2 \times$ VD
5	Adding or averaging step yields	multiply the fractions
6	Purity and yield on the same substance	purity to reactant, yield to product
7	"1 L of water $+$ solute" read as 1 L of solution	molality basis (1 kg of solvent)
8	Density in g L$^{-1}$ in a g mL$^{-1}$ formula	$1000d$ needs $d$ in g mL$^{-1}$
9	Water added $=$ final volume	added water $= V_2 - V_1$
10	Averaging molarities of unequal volumes	$(M_1V_1 + M_2V_2)/(V_1+V_2)$
11	Equal masses treated as arithmetic mean	convert to moles; harmonic mean
12	Mixing 22.4 L and 22.7 L	22.4 L at 1 atm (most JEE questions); 22.7 L at 1 bar
13	ppm per litre of concentrated solution	per $10^6$ g of solution; mg L$^{-1}$ only for dilute aqueous
14	Dropping the 2 in $\mathrm{Cr_2O_7^{2-}}$ or $\mathrm{C_2O_4^{2-}}$	per formula unit: 6 and 2

[JEE Main] Translate before computing: "VD 22" is $M = 44$, "acidified" is $n = 5$, "80% yield" is $\times 0.8$, "1.2 g mL$^{-1}$" is $1000d = 1200$.

Solved Examples

Question 1: n-factor and equivalent mass depend on the reaction

Find the equivalent mass of (a) $\mathrm{H_3PO_3}$ when completely neutralised by $\mathrm{NaOH}$, (b) $\mathrm{H_3PO_4}$ in the reaction $\mathrm{H_3PO_4 + 2NaOH \rightarrow Na_2HPO_4 + 2H_2O}$, and (c) $\mathrm{H_2SO_4}$ in $\mathrm{H_2SO_4 + KOH \rightarrow KHSO_4 + H_2O}$. (Molar masses: $\mathrm{H_3PO_3} = 82$, $\mathrm{H_3PO_4} = 98$, $\mathrm{H_2SO_4} = 98$ g mol$^{-1}$.)

Answer: (a) $\mathrm{H_3PO_3}$ has two $\mathrm{P-OH}$ and one $\mathrm{P-H}$. Only $\mathrm{O-H}$ hydrogens are replaceable, so $n = 2$. $$E = \frac{82}{2} = 41\ \mathrm{g\ eq^{-1}}$$ (b) In this reaction $\mathrm{H_3PO_4}$ gives up only 2 protons, so $n = 2$: $$E = \frac{98}{2} = 49\ \mathrm{g\ eq^{-1}}$$ (c) Only one proton is transferred, so $n = 1$: $$E = \frac{98}{1} = 98\ \mathrm{g\ eq^{-1}}$$

Ans: (a) 41, (b) 49, (c) 98 g eq$^{-1}$.

Watch out: The n-factor belongs to the reaction, not the formula. $\mathrm{H_3PO_4}$ can have $n = 1$, 2 or 3 ($E = 98$, 49 or 32.7).

Question 2: Equivalent mass of $\mathrm{KMnO_4}$ and $\mathrm{K_2Cr_2O_7}$ in different media

Calculate the equivalent mass of $\mathrm{KMnO_4}$ ($M = 158$) in acidic, neutral and strongly alkaline media, and of $\mathrm{K_2Cr_2O_7}$ ($M = 294$) in acidic medium.

Answer: First the oxidation number of Mn in $\mathrm{MnO_4^-}$: $x + 4(-2) = -1$, so $x = +7$.

Acidic: $\mathrm{Mn^{+7} \rightarrow Mn^{2+}}$, change $= 5$. $E = 158/5 = 31.6$.

Neutral: $\mathrm{Mn^{+7} \rightarrow MnO_2}$ (Mn is $+4$), change $= 3$. $E = 158/3 = 52.7$.

Strongly alkaline: $\mathrm{MnO_4^- \rightarrow MnO_4^{2-}}$ (Mn is $+6$), change $= 1$. $E = 158/1 = 158$.

Dichromate: $2x + 7(-2) = -2$, so $x = +6$. Each Cr goes to $+3$ (change 3), and there are two Cr per formula unit, so $n = 6$. $$E = \frac{294}{6} = 49\ \mathrm{g\ eq^{-1}}$$

Ans: $\mathrm{KMnO_4}$: 31.6 (acidic), 52.7 (neutral), 158 (alkaline); $\mathrm{K_2Cr_2O_7}$: 49 g eq$^{-1}$.

Watch out: $n$ is the total change per formula unit; the two chromiums are where a factor of 2 gets lost.

Question 3: A titration with the law of equivalents — no balanced equation required

25.0 mL of a $\mathrm{Na_2CO_3}$ solution requires 20.0 mL of 0.15 M $\mathrm{H_2SO_4}$ for complete neutralisation. Find (a) the normality and molarity of the $\mathrm{Na_2CO_3}$ solution and (b) the mass of $\mathrm{Na_2CO_3}$ ($M = 106$) present per litre.

Answer: $\mathrm{H_2SO_4}$ is dibasic, so its normality is $2 \times 0.15 = 0.30$ N.

By the law of equivalents, $N_1 V_1 = N_2 V_2$: $$N_{\mathrm{Na_2CO_3}} \times 25.0 = 0.30 \times 20.0 \quad \Rightarrow \quad N_{\mathrm{Na_2CO_3}} = \frac{6.0}{25.0} = 0.24\ \text{N}$$

For $\mathrm{Na_2CO_3}$ the n-factor is 2 (cationic charge $2 \times (+1)$), so $M = N/n = 0.24/2 = 0.12$ M.

Mass per litre: $M \times M_2 = 0.12 \times 106 = 12.72$ g L$^{-1}$, or $N \times E = 0.24 \times 53 = 12.72$ g L$^{-1}$.

Ans: (a) 0.24 N, 0.12 M; (b) 12.72 g L$^{-1}$.

Question 4: Equivalent mass of a metal from the hydrogen it liberates

0.90 g of a metal reacts with excess dilute $\mathrm{HCl}$ to liberate 1.12 L of $\mathrm{H_2}$ at STP (1 atm). Find the equivalent mass of the metal. If the metal is trivalent, identify it.

Answer: One mole of $\mathrm{H_2}$ (22.4 L at 1 atm STP) is 2 equivalents, so 1 equivalent is 11.2 L: $$\text{eq of } \mathrm{H_2} = \frac{1.12}{11.2} = 0.100$$

Equivalents of metal $=$ equivalents of $\mathrm{H_2} = 0.100$, so $$E = \frac{\text{mass}}{\text{equivalents}} = \frac{0.90}{0.100} = 9.0\ \mathrm{g\ eq^{-1}}$$

Trivalent: atomic mass $= E \times n = 9.0 \times 3 = 27$, aluminium.

Ans: $E = 9.0$ g eq$^{-1}$; the metal is Al (atomic mass 27).

Watch out: 1 equivalent of any substance liberates or consumes 11.2 L of $\mathrm{H_2}$ at 1 atm STP (or 5.6 L of $\mathrm{O_2}$, $n = 4$).

Question 5: Vapour density — molar mass, and the degree of dissociation of $\mathrm{PCl_5}$

(a) A gaseous compound has vapour density 32. What is its molar mass, and what volume does 6.4 g of it occupy at STP (1 atm)? (b) The vapour density of a $\mathrm{PCl_5}$ sample at a certain temperature is found to be 62. Calculate its degree of dissociation. ($M_{\mathrm{PCl_5}} = 208.5$.)

Answer: (a) $M = 2 \times \text{VD} = 2 \times 32 = 64$ g mol$^{-1}$ ($\mathrm{SO_2}$). Moles $= 6.4/64 = 0.10$, volume $= 0.10 \times 22.4 = 2.24$ L.

(b) Theoretical VD: $D = 208.5/2 = 104.25$. In $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$, $n = 2$, so $$\alpha = \frac{D - d}{(n - 1)d} = \frac{104.25 - 62}{62} = \frac{42.25}{62} = 0.681$$

Ans: (a) $M = 64$ g mol$^{-1}$, volume $= 2.24$ L; (b) $\alpha \approx 0.68$, about 68% dissociated.

Watch out: If $\alpha$ comes out negative, $D$ and $d$ are swapped. At 1 bar STP (22.7 L) the volume in (a) would be 2.27 L.

Question 6: Sequential reactions — overall yield and the mass you must start with

Benzene is converted to a product P in three steps with yields of 80%, 75% and 60%, each step being 1 : 1 in moles. (a) What is the overall yield? (b) What mass of benzene ($M = 78$) must be taken to obtain 36.0 g of P ($M = 200$)?

Answer: (a) Yields multiply: $$0.80 \times 0.75 \times 0.60 = 0.36 = 36\%$$

(b) Moles of P $= 36.0/200 = 0.180$ mol. Only 36% of the benzene becomes P, so $$n_{\text{benzene}} = \frac{0.180}{0.36} = 0.500\ \text{mol}$$ Mass of benzene $= 0.500 \times 78 = 39.0$ g.

Ans: (a) 36%; (b) 39.0 g of benzene.

Watch out: Going backwards, divide by the yield fraction. Multiplying gives $0.180 \times 0.36 \times 78 = 5.1$ g, too little even at 100% yield.

Question 7: Per cent purity from the volume of gas evolved

A 4.00 g sample of impure zinc reacts completely with excess dilute $\mathrm{H_2SO_4}$ to give 1.12 L of $\mathrm{H_2}$ at STP (1 atm). The impurities are inert. Find the per cent purity of the zinc. ($\mathrm{Zn} = 65.4$.)

Answer: $\mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2}$: 1 mol Zn gives 1 mol $\mathrm{H_2}$.

Moles of $\mathrm{H_2} = 1.12/22.4 = 0.0500$, so pure zinc $= 0.0500 \times 65.4 = 3.27$ g. $$\%\ \text{purity} = \frac{3.27}{4.00} \times 100 = 81.75\% \approx 81.8\%$$

By equivalents: eq of $\mathrm{H_2} = 1.12/11.2 = 0.100$, Zn $= 0.100 \times 32.7 = 3.27$ g.

Ans: The sample is about 81.8% pure zinc.

Question 8: POAC on an ugly reaction [Advanced]

Iron pyrites, $\mathrm{FeS_2}$ ($M = 120$), is roasted in air: $\mathrm{FeS_2 + O_2 \rightarrow Fe_2O_3 + SO_2}$ (unbalanced). From 60.0 g of $\mathrm{FeS_2}$, find (a) the mass of $\mathrm{Fe_2O_3}$ ($M = 160$) formed, (b) the volume of $\mathrm{SO_2}$ at STP (1 atm), and (c) the mass of $\mathrm{O_2}$ consumed — without balancing the equation.

Answer: Moles of $\mathrm{FeS_2} = 60.0/120 = 0.500$ mol.

(a) POAC on iron; $\mathrm{Fe_2O_3}$ has 2 Fe per formula unit: $$1 \times n_{\mathrm{FeS_2}} = 2 \times n_{\mathrm{Fe_2O_3}} \Rightarrow n_{\mathrm{Fe_2O_3}} = 0.250\ \text{mol} = 0.250 \times 160 = 40.0\ \text{g}$$

(b) POAC on sulphur: $$2 \times n_{\mathrm{FeS_2}} = 1 \times n_{\mathrm{SO_2}} \Rightarrow n_{\mathrm{SO_2}} = 1.00\ \text{mol} = 22.4\ \text{L}$$

(c) POAC on oxygen, now that both products are known: O atoms $= 3 \times 0.250 + 2 \times 1.00 = 0.75 + 2.00 = 2.75$ mol. Each $\mathrm{O_2}$ supplies 2: $$n_{\mathrm{O_2}} = \frac{2.75}{2} = 1.375\ \text{mol} = 1.375 \times 32 = 44.0\ \text{g}$$

Mass check: reactants $60.0 + 44.0 = 104.0$ g; products $40.0 + 1.00 \times 64 = 104.0$ g.

Ans: (a) 40.0 g $\mathrm{Fe_2O_3}$; (b) 22.4 L $\mathrm{SO_2}$; (c) 44.0 g $\mathrm{O_2}$.

Watch out: Oxygen is split between two products, so balance Fe and S first. (Balanced: $\mathrm{4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2}$.)

Question 9: Mass per cent to molarity, molality and mole fraction

An aqueous solution of $\mathrm{H_2SO_4}$ is 29.4% by mass and has a density of 1.20 g mL$^{-1}$. Calculate its molarity, molality and the mole fraction of $\mathrm{H_2SO_4}$. ($M_{\mathrm{H_2SO_4}} = 98$, $M_{\mathrm{H_2O}} = 18$.)

Answer: Basis 100 g of solution: 29.4 g of $\mathrm{H_2SO_4}$ ($29.4/98 = 0.300$ mol) and 70.6 g of water ($70.6/18 = 3.922$ mol).

Volume $= 100/1.20 = 83.33$ mL $= 0.08333$ L: $$M = \frac{0.300}{0.08333} = 3.60\ \text{M}$$ (Check: $10dw/M_2 = 10 \times 1.20 \times 29.4/98 = 3.60$.)

Solvent $= 0.0706$ kg: $$m = \frac{0.300}{0.0706} = 4.25\ \text{mol kg}^{-1}$$

$$x_{\mathrm{H_2SO_4}} = \frac{0.300}{0.300 + 3.922} = \frac{0.300}{4.222} = 0.0711$$

Ans: $M = 3.60$ M, $m = 4.25$ m, $x_{\mathrm{H_2SO_4}} = 0.071$.

Watch out: $m \approx M$ holds only for dilute aqueous solutions; here 3.60 and 4.25 differ a lot.

Question 10: Molarity to molality with the derived formula, and back

(a) A 2.00 M $\mathrm{NaOH}$ solution has density 1.08 g mL$^{-1}$. Find its molality. (b) A 3.00 molal aqueous solution of $\mathrm{NaCl}$ ($M = 58.5$) has density 1.10 g mL$^{-1}$. Find its molarity.

Answer: (a) Basis 1 L of solution: mass $= 1000 \times 1.08 = 1080$ g, $\mathrm{NaOH} = 2.00 \times 40 = 80$ g, water $= 1080 - 80 = 1000$ g $= 1.000$ kg. $$m = \frac{2.00}{1.000} = 2.00\ \text{mol kg}^{-1}$$ Check: $\dfrac{1000M}{1000d - MM_2} = \dfrac{2000}{1080 - 80} = 2.00$.

(b) Basis 1 kg of solvent: $\mathrm{NaCl} = 3.00 \times 58.5 = 175.5$ g, total mass $1175.5$ g, volume $= 1175.5/1.10 = 1068.6$ mL $= 1.0686$ L. $$M = \frac{3.00}{1.0686} = 2.81\ \text{M}$$ Check: $\dfrac{1000dm}{1000 + mM_2} = \dfrac{3300}{1175.5} = 2.81$.

Ans: (a) 2.00 m; (b) 2.81 M.

Watch out: In (a) $m = M$ only because the solvent mass came out as exactly 1 kg; that is not a rule.

Question 11: Mixing an acid with a base — the resultant normality

100 mL of 0.30 M $\mathrm{H_2SO_4}$ is mixed with 200 mL of 0.20 M $\mathrm{NaOH}$. Assuming additive volumes, find (a) the normality and nature of the resulting solution and (b) the molar concentration of the ion in excess.

Answer: Acid: $N = 2 \times 0.30 = 0.60$ N; equivalents $= 0.60 \times 0.100 = 0.0600$.

Base: $N = 0.20$ N; equivalents $= 0.20 \times 0.200 = 0.0400$.

Acid is in excess by $0.0600 - 0.0400 = 0.0200$ eq, so the solution is acidic.

(a) Total volume $= 300$ mL $= 0.300$ L: $$N = \frac{0.0200}{0.300} = 0.0667\ \text{N}$$

(b) For $\mathrm{H^+}$, one equivalent is one mole, so $\mathrm{[H^+]} = 0.0667$ M $\approx 0.067$ M. (Unreacted $\mathrm{H_2SO_4}$: $0.0100$ mol in 0.300 L $= 0.0333$ M; $2 \times 0.0333 = 0.0667$.)

Ans: (a) 0.067 N, acidic; (b) $\mathrm{[H^+]} \approx 0.067$ M.

Watch out: By moles (0.030 vs 0.040) the base looks in excess; by equivalents (0.060 vs 0.040) the acid is.

Question 12: Average molar mass of a gas mixture and a ppm conversion

(a) A mixture of $\mathrm{CH_4}$ and $\mathrm{C_2H_6}$ has vapour density 11.0. Find the mole fraction of methane. (b) A 500 mL sample of drinking water (density 1.00 g mL$^{-1}$) contains 0.75 mg of fluoride ion. Express the fluoride concentration in ppm and in ppb.

Answer: (a) $M_{\text{avg}} = 2 \times 11.0 = 22.0$ g mol$^{-1}$. With mole fraction $x$ of $\mathrm{CH_4}$: $$16x + 30(1 - x) = 22.0 \Rightarrow 30 - 14x = 22.0 \Rightarrow x = \frac{8.0}{14} = 0.571$$

(b) Mass of water $= 500$ g. Mass of fluoride $= 0.75$ mg $= 7.5 \times 10^{-4}$ g. $$\text{ppm} = \frac{7.5 \times 10^{-4}}{500} \times 10^{6} = 1.5\ \text{ppm}$$ ppb $= 1.5 \times 1000 = 1500$. (Check: 0.75 mg in 0.5 L is 1.5 mg L$^{-1}$ $\approx 1.5$ ppm.)

Ans: (a) $x_{\mathrm{CH_4}} = 0.571$ (about 4/7); (b) 1.5 ppm $= 1500$ ppb.

Watch out: ppm divides by the mass of solution; mg L$^{-1} \approx$ ppm only because $d = 1$ g mL$^{-1}$.