CUET 2025 Chemistry Question Paper with Answers & Solutions

Correct answer: A

Acetic acid is a weak acid that partially dissociates in water: $\ce{CH3COOH <=> CH3COO^- + H^+}$. Partial dissociation gives $i$ between 1 (no dissociation) and 2 (complete dissociation into two ions).

Correct answer: C

Molarity = (1 g / M) / 1 L = 1/M. Lower molar mass gives higher molarity. M values: NaOH 40, NaCl 58.5, KCl 74.5, Glucose 180. Decreasing molarity: NaOH(B) > NaCl(C) > KCl(D) > Glucose(A), i.e. (B),(C),(D),(A).

Correct answer: D

Saturated = max solute (III); Isotonic = same osmotic pressure (IV); Binary = two components (I); Hypertonic = higher osmotic pressure (II). So (A)-(III),(B)-(IV),(C)-(I),(D)-(II).

Correct answer: C

By Henry's law, solubility of a gas is proportional to its partial pressure. At high altitude atmospheric (and hence oxygen partial) pressure is low, so less O2 dissolves in blood, causing anoxia.

Correct answer: A

In 100 g solution: 20 g KI, 80 g water. Moles KI = 20/166 = 0.1205 mol. Molality = 0.1205 mol / 0.080 kg = 1.506 mol kg$^{-1}$ ≈ 1.5. (Density not needed for molality.)

Correct answer: C

$E_{cell}$ is the cell potential (electromotive force), measured in volts (V).

Correct answer: A

Cell constant = cm$^{-1}$ (I); Molar conductance = ohm$^{-1}$ cm$^2$ mol$^{-1}$ (II); Specific conductance (conductivity) = ohm$^{-1}$ cm$^{-1}$ (III); Conductance = ohm$^{-1}$ (IV). So (A)-(I),(B)-(II),(C)-(III),(D)-(IV).

Correct answer: D

Mercury cell is a primary (non-rechargeable) cell, not a combustion device. Correct facts: net reaction $\ce{Zn(Hg) + HgO(s) -> ZnO(s) + Hg(l)}$ (C) and it is a low-current device used in hearing aids/watches (D). So (C) and (D) only.

Correct answer: C

The lead storage (lead-acid) battery is the rechargeable secondary cell used in automobiles and inverters.

Correct answer: C

When an external potential greater than $E_{cell}$ is applied ($E_{ext} > E_{cell}$), current reverses and the galvanic cell functions as an electrolytic cell.

Correct answer: C

Desferrioxime B (D-penicillamine/desferrioxime) is used in treatment of iron poisoning (excess iron), not lead poisoning. The other statements are correct.

Correct answer: C

$\ce{Ti^3+}$ = $3d^1$; $\ce{Cr^2+}$ = $3d^4$; $\ce{Mn^+}$ = $3d^5 4s^1$ (5 d-electrons); $\ce{Cu^+}$ = $3d^{10}$. Increasing 3d electrons: Ti(1) < Mn(5) < Cr... wait check: Ti=1, Cr=4, Mn+=5(d5 4s1), Cu+=10. So order Ti(C) < Cr(A) < Mn(D) < Cu(B). That is (C),(A),(D),(B) = option B. Re-evaluating Mn+: Mn(Z=25)=$3d^5 4s^2$, Mn+ removes one 4s = $3d^5 4s^1$, so 3d=5; Cr2+ = $3d^4$=4. Thus Ti(1)<Cr(4)<Mn(5)<Cu(10) = (C),(A),(D),(B). Option B.

Correct answer: A

Lanthanum (Z=57) has configuration $[Xe]5d^1 6s^2$ — the 4f subshell is empty in La.

Correct answer: B

Mg-based alloy (misch metal) used in bullets/shells (I); Lanthanoid oxide used in TV screens/phosphors (III); Mixed lanthanoid oxides used as catalysts in petroleum cracking (II); Misch metal = lanthanoid metal (~95%) + iron (~5%) (IV). So (A)-(I),(B)-(III),(C)-(II),(D)-(IV).

Correct answer: A

Hydrated $\ce{Fe^3+}$ ion (as $\ce{[Fe(H2O)6]^3+}$, often hydrolysed) appears yellow/pale yellow-brown in aqueous solution.

Correct answer: B

$\ce{KMnO4}$ oxidises HCl to chlorine ($\ce{Cl2}$), consuming the oxidant and interfering with the reaction, so dilute $\ce{H2SO4}$ is used instead.

Correct answer: C

Acidified $\ce{K2Cr2O7}$ oxidises sulphide ions ($\ce{S^2-}$, oxidation state −2) to elemental sulphur (S, state 0): $\ce{3H2S + Cr2O7^2- + 8H^+ -> 2Cr^3+ + 3S + 7H2O}$.

Correct answer: C

Spectrochemical series (decreasing field strength): $\ce{CN^-}$ > en > $\ce{NCS^-}$ (N-bonded) > $\ce{S^2-}$. So (D),(B),(C),(A).

Correct answer: B

Ligands alphabetical (ammine before chlorido). Oxidation state of Pt: x + 0 × 2 (NH3 neutral) + (−1) × 2 = 0, so x = +2. Name: diamminedichloridoplatinum(II).

Correct answer: A

Fe is +3 ($d^5$); $\ce{CN^-}$ is strong field, so low-spin with one unpaired electron: $d^2sp^3$ hybridization, inner orbital octahedral, paramagnetic with μ ≈ 1.73 BM (not 5.92). Correct statements: (A) paramagnetic and (D) $d^2sp^3$.

Correct answer: A

Ambident nucleophiles = cyanides and nitrites (III); Plane polarized light produced by Nicol prism (IV); Superimposable mirror image = symmetrical (achiral) object (I); β-elimination follows Saytzeff rule (II). So (A)-(III),(B)-(IV),(C)-(I),(D)-(II).

Correct answer: C

SN2 proceeds via backside attack giving Walden inversion — inversion of configuration.

Correct answer: C

Gabriel phthalimide synthesis produces primary aliphatic amines (aromatic amines cannot be prepared as aryl halides do not react).

Correct answer: C

Coupling of an aryl halide with an alkyl halide using sodium in dry ether to give an alkyl-substituted arene is the Wurtz–Fittig reaction.

Correct answer: C

Hydroboration-oxidation: alkene + $\ce{B2H6}$ (diborane) gives trialkylborane, then oxidation with $\ce{H2O2}$ in $\ce{OH^-}$ (alkaline), with $\ce{H2O}$ involved. All four reagents are used.

Correct answer: A

Starch is the main storage polysaccharide in plants (glycogen is the animal storage form, cellulose is structural).

Correct answer: D

The Hell-Volhard-Zelinsky reaction halogenates carboxylic acids at the α-carbon (using $\ce{X2}$ and red P) to give α-halocarboxylic acids.

Correct answer: B

Nucleophilic addition reactivity: aldehydes > ketones, and more/larger alkyl groups decrease reactivity (steric + electronic). Order of increasing reactivity: butanone(D) < propanone(B) < propanal(C) < ethanal(A). So (D),(B),(C),(A).

Correct answer: C

Benzene → bromobenzene ($\ce{Br2/FeBr3}$) → Grignard (Mg, dry ether) → benzoic acid ($\ce{CO2}$, then $\ce{H3O^+}$) → methyl benzoate (Fischer esterification: methanol, conc. $\ce{H2SO4}$). All four reagents required.

Correct answer: D

Acetone ($\ce{CH3COCH3}$) has a $\ce{CH3CO-}$ group and gives a positive iodoform ($\ce{I2/NaOH}$) test (yellow precipitate); benzophenone ($\ce{C6H5COC6H5}$) has no methyl ketone group and does not. Both give 2,4-DNP; neither reduces Fehling/Tollens (they are ketones).

Correct answer: B

Aldol condensation requires at least one α-hydrogen so that an enolate (carbanion) can form.

Correct answer: B

N in trimethylamine is $sp^3$ with three bond pairs and one lone pair, giving a pyramidal (trigonal pyramidal) geometry.

Correct answer: A

In cleavage of alkyl aryl ethers by HX, the aryl-O bond is not cleaved; the alkyl-O bond breaks to give phenol and alkyl halide: $\ce{C6H5-O-R + HX -> C6H5OH + RX}$.

Correct answer: B

The electrophile (nitrating species) in benzene nitration is the nitronium ion $\ce{NO2^+}$, generated by $\ce{HNO3 + 2H2SO4 -> NO2^+ + H3O^+ + 2HSO4^-}$.

Correct answer: A

Azo coupling needs an electron-rich aromatic (activated by –OH or –NH2). Nitrobenzene has the strongly deactivating –NO2 group and will not couple. Aniline, o-toluidine and phenol all couple.

Correct answer: B

Starch consists of amylose (water-soluble, about 15–20%) and amylopectin (insoluble, about 80–85%) per NCERT.

Correct answer: B

Globular proteins (chains coiled into spherical shape) include insulin and albumin. Keratin and myosin are fibrous proteins. So (A) and (C) only.

Correct answer: C

The helical/pleated secondary structures of proteins (α-helix and β-sheet) are stabilised by hydrogen bonds between C=O and N–H groups.

Correct answer: B

One-letter codes: Lysine = K (III); Tryptophan = W (I); Tyrosine = Y (IV); Glutamine = Q (II). So (A)-(III),(B)-(I),(C)-(IV),(D)-(II).

Correct answer: C

Electron-withdrawing nitro groups increase acidity; more nitro groups = more acidic. Increasing order: phenol(D) < 3-nitrophenol(A) < 3,5-dinitrophenol(B) < 2,4,6-trinitrophenol/picric acid(C). So (D),(A),(B),(C).

Correct answer: C

Reducing volume to 1/4 increases each concentration 4 times. Rate = k[A][B], so new rate = k(4[A])(4[B]) = 16 k[A][B] = 16 times the initial rate.

Correct answer: A

Overall order = x + y = 1/2 + 3/2 = 2, i.e. second order.

Correct answer: C

k = rate / [P]^{3/2}. Units = (bar min$^{-1}$)/(bar$^{3/2}$) = bar$^{1-3/2}$ min$^{-1}$ = bar$^{-1/2}$ min$^{-1}$.

Correct answer: C

Rate ∝ [A]^n. $3^n = 27 = 3^3$, so n = 3. Third order.

Correct answer: C

A catalyst provides an alternate pathway of lower activation energy; it does not change ΔG, ΔH, or the equilibrium constant.

Correct answer: D

At higher temperature (443 K) ethanol undergoes intramolecular dehydration to ethene (elimination); at lower temperature (413 K) intermolecular dehydration gives ethoxyethane (diethyl ether).

Correct answer: B

The methoxy group is o/p-directing; due to steric hindrance the para product predominates, so p-bromoanisole is the major product.

Correct answer: A

Williamson ether synthesis proceeds by SN2; the alkoxide is a strong nucleophile and works best with primary alkyl halides.

Correct answer: D

Reactivity of HX towards ether cleavage: HI > HBr > HCl. HI is most reactive because iodide is the best nucleophile and HI is the strongest acid.

Correct answer: C

Anisole is $\ce{C6H5-O-CH3}$ (methoxybenzene), an aryl alkyl (phenyl alkyl) ether with one aryl and one alkyl group on oxygen.