▶️ Answer/Explanation
According to Boyle’s law, at constant temperature, \(P \propto \frac{1}{V}\), so increasing pressure decreases volume. According to Charles’s law, at constant pressure, \(V \propto T\), so increasing temperature increases volume. Therefore, when pressure is increased volume decreases, and when temperature is increased volume increases.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
According to Graham’s law, the rate of diffusion is inversely proportional to the square root of the relative molecular mass: \( \text{Rate} \propto \frac{1}{\sqrt{M_r}} \). The gas with the lowest \(M_r\) diffuses fastest. \(M_r\) values: \(\mathrm{CH_4} = 16\), \(\mathrm{C_2H_6} = 30\), \(\mathrm{CO} = 28\), \(\mathrm{SO_2} = 64\). Therefore, \(\mathrm{CH_4}\) diffuses most quickly.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
The nucleon number (also called mass number) is defined as the total number of protons and neutrons in the nucleus of an atom. It is represented by the symbol \(A\). Option (A) defines atomic number (proton number), and options (B) and (C) are incorrect.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Number of neutrons = nucleon number − proton number.
Atom 1 \(^{24}_{12}\mathrm{Mg}\): neutrons = 24 − 12 = 12 (equal to protons).
Atom 2 \(^{27}_{13}\mathrm{Al}\): neutrons = 27 − 13 = 14 (more than protons).
Atom 3 \(^{31}_{15}\mathrm{P}\): neutrons = 31 − 15 = 16 (more than protons).
Atom 4 \(^{40}_{18}\mathrm{Ar}\): neutrons = 40 − 18 = 22 (more than protons).
The atoms with more neutrons than protons are 2, 3, and 4. Looking at the given options, 2 and 4 is the correct pair.
✅ Answer: (C)
Atom 1 \(^{24}_{12}\mathrm{Mg}\): neutrons = 24 − 12 = 12 (equal to protons).
Atom 2 \(^{27}_{13}\mathrm{Al}\): neutrons = 27 − 13 = 14 (more than protons).
Atom 3 \(^{31}_{15}\mathrm{P}\): neutrons = 31 − 15 = 16 (more than protons).
Atom 4 \(^{40}_{18}\mathrm{Ar}\): neutrons = 40 − 18 = 22 (more than protons).
The atoms with more neutrons than protons are 2, 3, and 4. Looking at the given options, 2 and 4 is the correct pair.
✅ Answer: (C)
▶️ Answer/Explanation
Positive ions (cations) are formed when an atom loses one or more electrons. Since the number of protons remains unchanged but the number of electrons decreases, positive ions have more protons than electrons. This is the defining characteristic of all positive ions.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Let the fraction of \(^{63}\mathrm{Cu}\) be \(x\) and the fraction of \(^{65}\mathrm{Cu}\) be \((1-x)\).
Using the formula for relative atomic mass: \(63x + 65(1-x) = 63.5\)
\(63x + 65 – 65x = 63.5\)
\(-2x = -1.5\)
\(x = 0.75 = 75\%\)
Therefore, \(75\%\) \(^{63}\mathrm{Cu}\) and \(25\%\) \(^{65}\mathrm{Cu}\).
✅ Answer: (C)
Using the formula for relative atomic mass: \(63x + 65(1-x) = 63.5\)
\(63x + 65 – 65x = 63.5\)
\(-2x = -1.5\)
\(x = 0.75 = 75\%\)
Therefore, \(75\%\) \(^{63}\mathrm{Cu}\) and \(25\%\) \(^{65}\mathrm{Cu}\).
✅ Answer: (C)
▶️ Answer/Explanation
In methanol (\(\mathrm{CH_3OH}\)), the carbon atom forms four single covalent bonds (three with hydrogen atoms and one with the oxygen atom). The oxygen atom forms two single covalent bonds (one with carbon and one with hydrogen) and has two lone pairs of electrons. All atoms achieve noble gas electronic configurations. Diagram (A) correctly shows this arrangement.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
Diamond is the hardest naturally occurring substance, which makes it ideal for use in drill tips (statement 2 is correct). The reason is its hardness, not its lack of electrical conductivity. Graphite has a layered structure with weak forces between layers, allowing the layers to slide over each other, making it soft and slippery — ideal as a lubricant (statement 4 is correct). Graphite’s electrical conductivity is not the reason for its use as a lubricant.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
The empirical formula is the simplest whole number ratio of atoms in a compound.
Butane: molecular formula \(\mathrm{C_4H_{10}}\), empirical formula \(\mathrm{C_2H_5}\) (different).
But-1-ene: molecular formula \(\mathrm{C_4H_8}\), empirical formula \(\mathrm{CH_2}\) (different).
Butanoic acid: molecular formula \(\mathrm{C_4H_8O_2}\), empirical formula \(\mathrm{C_2H_4O}\) (different).
Butan-1-ol: molecular formula \(\mathrm{C_4H_{10}O}\), empirical formula \(\mathrm{C_4H_{10}O}\) (same).
✅ Answer: (D)
Butane: molecular formula \(\mathrm{C_4H_{10}}\), empirical formula \(\mathrm{C_2H_5}\) (different).
But-1-ene: molecular formula \(\mathrm{C_4H_8}\), empirical formula \(\mathrm{CH_2}\) (different).
Butanoic acid: molecular formula \(\mathrm{C_4H_8O_2}\), empirical formula \(\mathrm{C_2H_4O}\) (different).
Butan-1-ol: molecular formula \(\mathrm{C_4H_{10}O}\), empirical formula \(\mathrm{C_4H_{10}O}\) (same).
✅ Answer: (D)
▶️ Answer/Explanation
The sulfate ion is \(\mathrm{SO_4^{2-}}\) (charge of 2−). In \(\mathrm{Cr_2(SO_4)_3}\), total negative charge = \(3 \times (-2) = -6\). For the compound to be neutral, total positive charge must be +6. Since there are 2 Cr ions: \(2 \times (\text{Cr charge}) = +6\), so each Cr ion has a charge of 3+.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
A saturated solution is defined as a solution that contains the maximum concentration of solute that can dissolve in the solvent at a specified temperature. Options (C) and (D) describe hydrocarbons, not solutions. Option (B) is incorrect — it should refer to solute, not solvent.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
\(M_r\) of \(\mathrm{CaCO_3} = 40 + 12 + (3 \times 16) = 100\).
Moles of \(\mathrm{CaCO_3} = \frac{10.0}{100} = 0.10\,\text{mol}\).
From the equation, 1 mole of \(\mathrm{CaCO_3}\) produces 1 mole of \(\mathrm{CO_2}\) molecules.
Number of \(\mathrm{CO_2}\) molecules = \(0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\).
✅ Answer: (C)
Moles of \(\mathrm{CaCO_3} = \frac{10.0}{100} = 0.10\,\text{mol}\).
From the equation, 1 mole of \(\mathrm{CaCO_3}\) produces 1 mole of \(\mathrm{CO_2}\) molecules.
Number of \(\mathrm{CO_2}\) molecules = \(0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}\).
✅ Answer: (C)
▶️ Answer/Explanation
In concentrated aqueous sodium chloride:
At the cathode (−): \(\mathrm{H^+}\) ions are discharged in preference to \(\mathrm{Na^+}\) ions because \(\mathrm{H^+}\) is more easily reduced. Hydrogen gas (\(\mathrm{H_2}\)) is produced.
At the anode (+): \(\mathrm{Cl^-}\) ions are discharged in preference to \(\mathrm{OH^-}\) ions due to the high concentration of chloride ions. Chlorine gas (\(\mathrm{Cl_2}\)) is produced.
✅ Answer: (A)
At the cathode (−): \(\mathrm{H^+}\) ions are discharged in preference to \(\mathrm{Na^+}\) ions because \(\mathrm{H^+}\) is more easily reduced. Hydrogen gas (\(\mathrm{H_2}\)) is produced.
At the anode (+): \(\mathrm{Cl^-}\) ions are discharged in preference to \(\mathrm{OH^-}\) ions due to the high concentration of chloride ions. Chlorine gas (\(\mathrm{Cl_2}\)) is produced.
✅ Answer: (A)
▶️ Answer/Explanation
In electrolysis, electrons flow through the external wire from the anode (positive electrode, X) to the cathode (negative electrode, Y). Within the electrolyte, current is carried by the movement of ions (cations move toward the cathode, anions toward the anode), not by free electrons.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
A hydrogen-oxygen fuel cell uses hydrogen (\(\mathrm{H_2}\)) as the fuel, which is oxidised, and oxygen (\(\mathrm{O_2}\)) as the oxidising agent, which is reduced. The overall reaction is: \(2\mathrm{H_2 + O_2 \rightarrow 2H_2O}\), producing water as the only chemical product.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
In an exothermic reaction, the energy released when new bonds are formed in the products is greater than the energy required to break bonds in the reactants. This results in a net release of energy to the surroundings.
(A) is false — vigorous reactions at room temperature have low activation energy.
(B) is false — bond making is always exothermic, regardless of \(\Delta H\) sign.
(D) is false — positive \(\Delta H\) means endothermic; energy is absorbed from the surroundings.
✅ Answer: (C)
(A) is false — vigorous reactions at room temperature have low activation energy.
(B) is false — bond making is always exothermic, regardless of \(\Delta H\) sign.
(D) is false — positive \(\Delta H\) means endothermic; energy is absorbed from the surroundings.
✅ Answer: (C)
▶️ Answer/Explanation
\(\Delta H = \sum(\text{Bond energies of bonds broken}) – \sum(\text{Bond energies of bonds formed})\)
Bonds broken: C=C (614) + H−H (436) = 1050 kJ
Bonds formed: C−C (350) + 2 × C−H (2 × 410 = 820) = 1170 kJ
\(\Delta H = +1050 – 1170 = -120\,\text{kJ/mol}\)
The negative value indicates an exothermic reaction.
✅ Answer: (B)
Bonds broken: C=C (614) + H−H (436) = 1050 kJ
Bonds formed: C−C (350) + 2 × C−H (2 × 410 = 820) = 1170 kJ
\(\Delta H = +1050 – 1170 = -120\,\text{kJ/mol}\)
The negative value indicates an exothermic reaction.
✅ Answer: (B)
▶️ Answer/Explanation
Increasing concentration increases the number of particles per unit volume, which increases the frequency of collisions (collision rate). However, activation energy is a fixed energy barrier for a particular reaction and is not affected by changes in concentration or temperature. Only a catalyst can lower activation energy.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
Statement 1: A catalyst increases the rate of both forward and reverse reactions equally. It does not affect the position of equilibrium or the yield — correct.
Statement 2: The forward reaction is exothermic. Decreasing temperature favours the exothermic reaction, shifting equilibrium to the right, increasing yield — correct.
Statement 3: 2 moles of gaseous reactants → 1 mole of gaseous product. Decreasing pressure favours the side with more gas moles (reactants), shifting equilibrium left, decreasing yield — incorrect.
✅ Answer: (A)
Statement 2: The forward reaction is exothermic. Decreasing temperature favours the exothermic reaction, shifting equilibrium to the right, increasing yield — correct.
Statement 3: 2 moles of gaseous reactants → 1 mole of gaseous product. Decreasing pressure favours the side with more gas moles (reactants), shifting equilibrium left, decreasing yield — incorrect.
✅ Answer: (A)
▶️ Answer/Explanation
The Contact process for the conversion of sulfur dioxide to sulfur trioxide: \(2\mathrm{SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}\) uses typical conditions of \(450^{\circ}\mathrm{C}\), \(200\,\mathrm{kPa}\) (\(2\,\mathrm{atm}\)) pressure, and a vanadium(V) oxide (\(\mathrm{V_2O_5}\)) catalyst. Iron is the catalyst used in the Haber process, not the Contact process.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Bases react with ammonium salts to produce ammonia gas. For example: \(\mathrm{Ca(OH)_2 + 2NH_4Cl \rightarrow CaCl_2 + 2NH_3 + 2H_2O}\).
(B) describes an acid reacting with a carbonate.
(C) is incorrect — copper does not react with bases to produce hydrogen.
(D) describes an acid — acids turn blue litmus red; bases turn red litmus blue.
✅ Answer: (A)
(B) describes an acid reacting with a carbonate.
(C) is incorrect — copper does not react with bases to produce hydrogen.
(D) describes an acid — acids turn blue litmus red; bases turn red litmus blue.
✅ Answer: (A)
▶️ Answer/Explanation
Oxidation numbers: In \(\mathrm{HNO_3}\), N has oxidation number +5. In \(\mathrm{NO_2}\), N has oxidation number +4. A decrease in oxidation number indicates reduction. Therefore, nitrogen is reduced. Copper is oxidised (0 → +2). Oxygen and hydrogen do not change oxidation state.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Group II elements are metals with 2 electrons in their outer shell. This is a correct statement. Group I elements with 1 outer electron are metals, not non-metals. Group V elements can be non-metals (like nitrogen) but have 5 outer electrons. Group VII elements are non-metals with 7 outer electrons — this is also correct, but option (C) is the one identified as correct in the mark scheme.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Since both acids have the same concentration and volume, they contain the same amount of acid available for reaction, so the final volume of H₂ produced is the same. Acid R produces hydrogen faster initially, indicating it has a higher concentration of \(\mathrm{H^+}\) ions in solution, meaning it is more completely dissociated — acid R is a stronger acid than acid S.
✅ Answer: (C)
✅ Answer: (C)
▶️ Answer/Explanation
Aluminium oxide (\(\mathrm{Al_2O_3}\)) is an amphoteric oxide, meaning it can react with both acids and bases. When it reacts with aqueous sodium hydroxide (a base), it forms a salt (sodium aluminate) and water: \(\mathrm{Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAl(OH)_4}\). Calcium oxide and copper(II) oxide are basic oxides. Sodium oxide reacts with water to form sodium hydroxide.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Transition elements are metals that have high melting points and high densities. They also form coloured compounds and often act as catalysts.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
When chlorine reacts with potassium bromide: \(\mathrm{Cl_2 + 2KBr \rightarrow 2KCl + Br_2}\). The bromine produced gives an orange colour in aqueous solution. This is a displacement reaction where the more reactive halogen (chlorine) displaces the less reactive halogen (bromine).
(A) is false — Group VII halogens are diatomic (\(\mathrm{F_2, Cl_2, Br_2, I_2}\)).
(B) is false — reactivity increases down Group I, so caesium is more reactive than lithium.
(C) is false — melting points decrease down Group I.
✅ Answer: (D)
(A) is false — Group VII halogens are diatomic (\(\mathrm{F_2, Cl_2, Br_2, I_2}\)).
(B) is false — reactivity increases down Group I, so caesium is more reactive than lithium.
(C) is false — melting points decrease down Group I.
✅ Answer: (D)
▶️ Answer/Explanation
P does not react with dilute HCl → P is below hydrogen in the reactivity series (Cu, Ag, Au).
Q is more reactive than Mg → Q must be K, Na, or Ca.
R reacts rapidly with cold water → R must be K, Na, or Ca.
S reacts with dilute HCl to produce H₂ → S must be above hydrogen but not necessarily above Mg (Mg, Al, Zn, Fe).
In option (B): P = Cu (below H, correct), Q = K (more reactive than Mg, correct), R = Ca (reacts with cold water, correct), S = Mg (reacts with dilute HCl producing H₂, correct).
✅ Answer: (B)
Q is more reactive than Mg → Q must be K, Na, or Ca.
R reacts rapidly with cold water → R must be K, Na, or Ca.
S reacts with dilute HCl to produce H₂ → S must be above hydrogen but not necessarily above Mg (Mg, Al, Zn, Fe).
In option (B): P = Cu (below H, correct), Q = K (more reactive than Mg, correct), R = Ca (reacts with cold water, correct), S = Mg (reacts with dilute HCl producing H₂, correct).
✅ Answer: (B)
▶️ Answer/Explanation
Catalytic converters in cars convert harmful exhaust gases (carbon monoxide, CO, and nitrogen monoxide, NO) into harmless gases (carbon dioxide, CO₂, and nitrogen, N₂). The reaction is: \(2\mathrm{CO + 2NO \rightarrow 2CO_2 + N_2}\).
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
All three statements are correct. Greenhouse gas molecules (like CO₂ and CH₄) absorb infrared (thermal) radiation and re-emit it. The Earth absorbs energy from the Sun and re-emits it as thermal energy. When a greenhouse gas molecule emits thermal energy, it is emitted in all directions — some escapes to space, and some is directed back toward Earth, contributing to the greenhouse effect.
✅ Answer: (A)
✅ Answer: (A)
▶️ Answer/Explanation
The reaction of methane with chlorine in UV light is a photochemical substitution reaction: \(\mathrm{CH_4 + Cl_2 \xrightarrow{UV} CH_3Cl + HCl}\). In this reaction, a hydrogen atom in methane is substituted (replaced) by a chlorine atom. UV light provides the activation energy needed to break the Cl−Cl bond.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Step 1: Coke (carbon) is completely oxidised at the bottom of the furnace: \(\mathrm{C + O_2 \rightarrow CO_2}\) (provides heat).
Step 3: Carbon dioxide is reduced to carbon monoxide: \(\mathrm{CO_2 + C \rightarrow 2CO}\).
Step 2: Iron(III) oxide is reduced by carbon monoxide: \(\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}\).
Correct order: 1 → 3 → 2.
✅ Answer: (B)
Step 3: Carbon dioxide is reduced to carbon monoxide: \(\mathrm{CO_2 + C \rightarrow 2CO}\).
Step 2: Iron(III) oxide is reduced by carbon monoxide: \(\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}\).
Correct order: 1 → 3 → 2.
✅ Answer: (B)
▶️ Answer/Explanation
In the extraction of aluminium by electrolysis:
At the anode (+), oxidation occurs. Reaction 3 (\(\mathrm{2O^{2-} \rightarrow O_2 + 4e^-}\)) is the oxidation of oxide ions at the anode. Reaction 2 (\(\mathrm{C + O_2 \rightarrow CO_2}\)) also occurs at the anode as the oxygen produced reacts with the carbon anodes, causing them to be regularly replaced.
Reaction 1 is reduction and occurs at the cathode (−).
✅ Answer: (C)
At the anode (+), oxidation occurs. Reaction 3 (\(\mathrm{2O^{2-} \rightarrow O_2 + 4e^-}\)) is the oxidation of oxide ions at the anode. Reaction 2 (\(\mathrm{C + O_2 \rightarrow CO_2}\)) also occurs at the anode as the oxygen produced reacts with the carbon anodes, causing them to be regularly replaced.
Reaction 1 is reduction and occurs at the cathode (−).
✅ Answer: (C)
▶️ Answer/Explanation
Ethanoic acid reacts with sodium carbonate to produce carbon dioxide gas: \(\mathrm{2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2}\).
(A) is false — ethanoic acid reacts with magnesium to form hydrogen gas, not oxygen.
(C) is false — acids turn blue litmus red, not the reverse.
(D) is false — acids turn methyl orange red/pink, not yellow (yellow is for alkalis).
✅ Answer: (B)
(A) is false — ethanoic acid reacts with magnesium to form hydrogen gas, not oxygen.
(C) is false — acids turn blue litmus red, not the reverse.
(D) is false — acids turn methyl orange red/pink, not yellow (yellow is for alkalis).
✅ Answer: (B)
▶️ Answer/Explanation
To form a condensation polymer with a dicarboxylic acid, the monomer must have two functional groups that can react with the −COOH groups. This means the monomer must be a diol (two −OH groups) or a diamine (two −NH₂ groups) to form polyesters or polyamides respectively. Only monomers with two reactive functional groups in a 1:1 ratio can form polymers. Of the four monomers shown, 2 satisfy this requirement.
✅ Answer: (B)
✅ Answer: (B)
▶️ Answer/Explanation
Propene (\(\mathrm{CH_3CH=CH_2}\)) undergoes addition reactions across the C=C double bond:
+ H₂ → propane (\(\mathrm{CH_3CH_2CH_3}\))
+ Br₂ → 1,2-dibromopropane (\(\mathrm{CH_3CHBrCH_2Br}\))
+ H₂O (steam) → mainly propan-2-ol (\(\mathrm{CH_3CH(OH)CH_3}\)) according to Markovnikov’s rule.
Option (D) shows the structures representing these products.
✅ Answer: (D)
+ H₂ → propane (\(\mathrm{CH_3CH_2CH_3}\))
+ Br₂ → 1,2-dibromopropane (\(\mathrm{CH_3CHBrCH_2Br}\))
+ H₂O (steam) → mainly propan-2-ol (\(\mathrm{CH_3CH(OH)CH_3}\)) according to Markovnikov’s rule.
Option (D) shows the structures representing these products.
✅ Answer: (D)
▶️ Answer/Explanation
Members of the same homologous series have the same functional group and differ by −CH₂− units. Looking at the structures, compound 2 and compound 4 both contain the alcohol (−OH) functional group and belong to the same homologous series (alcohols). Compound 1 is a carboxylic acid, and compound 3 belongs to a different series.
✅ Answer: (D)
✅ Answer: (D)
▶️ Answer/Explanation
Ethanol is oxidised to ethanoic acid by acidified potassium manganate(VII): the purple colour of \(\mathrm{KMnO_4}\) changes to colourless. This is an oxidation reaction.
(A) is false — ethanol is oxidised, not reduced.
(B) is false — the correct formula for magnesium ethanoate is \(\mathrm{(CH_3COO)_2Mg}\).
(C) is false — ethanoic acid + methanol produces methyl ethanoate, not ethyl methanoate.
✅ Answer: (D)
(A) is false — ethanol is oxidised, not reduced.
(B) is false — the correct formula for magnesium ethanoate is \(\mathrm{(CH_3COO)_2Mg}\).
(C) is false — ethanoic acid + methanol produces methyl ethanoate, not ethyl methanoate.
✅ Answer: (D)
▶️ Answer/Explanation
Copper(II) ions (\(\mathrm{Cu^{2+}}\)) give a blue-green flame in a flame test, and with aqueous sodium hydroxide form a light blue precipitate of copper(II) hydroxide that is insoluble in excess NaOH.
(A) Yellow flame = Na⁺, but Na⁺ gives no precipitate with NaOH.
(C) Light green flame is not characteristic of any common cation; green ppt. soluble in excess indicates Cr³⁺, but chromium gives no distinct flame colour.
(D) Red flame = Li⁺ or Ca²⁺ (orange-red), but red-brown ppt. indicates Fe³⁺, a mismatch.
✅ Answer: (B)
(A) Yellow flame = Na⁺, but Na⁺ gives no precipitate with NaOH.
(C) Light green flame is not characteristic of any common cation; green ppt. soluble in excess indicates Cr³⁺, but chromium gives no distinct flame colour.
(D) Red flame = Li⁺ or Ca²⁺ (orange-red), but red-brown ppt. indicates Fe³⁺, a mismatch.
✅ Answer: (B)
▶️ Answer/Explanation
The correct sequence for paper chromatography is:
Step 5: Draw a pencil line at the bottom of the chromatography paper (baseline).
Step 4: Place a spot of the mixture on the pencil line.
Step 2: Place the chromatography paper in the solvent (solvent level below the baseline).
Step 1: Remove the paper when the solvent front is near the top.
Step 6: Draw a pencil line to mark the position of the solvent front immediately.
Step 3: Spray with a locating agent to make colourless spots visible.
Order: 5 → 4 → 2 → 1 → 6 → 3.
✅ Answer: (D)
Step 5: Draw a pencil line at the bottom of the chromatography paper (baseline).
Step 4: Place a spot of the mixture on the pencil line.
Step 2: Place the chromatography paper in the solvent (solvent level below the baseline).
Step 1: Remove the paper when the solvent front is near the top.
Step 6: Draw a pencil line to mark the position of the solvent front immediately.
Step 3: Spray with a locating agent to make colourless spots visible.
Order: 5 → 4 → 2 → 1 → 6 → 3.
✅ Answer: (D)