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Chemistry - 4 — MH-CET Full Length Paper MCQs with Solutions

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Sample questions with solutions

Q1 — easy
A compound forms a cubic close-packed (ccp) lattice of element $B$. Element $A$ occupies all the octahedral voids. The simplest formula of the compound is:
A. $AB$  ✓ Correct
B. $AB_2$
C. $A_2B$
D. $A_2B_3$
Solution: In a ccp lattice with $N$ atoms of $B$, the number of octahedral voids is exactly $N$. Since $A$ fills all octahedral voids, $A:B = 1:1$. Formula is $AB$.
Q2 — easy
The determination of molar mass of biomacromolecules such as proteins and polymers is most accurately carried out using which colligative property measurement?
A. Elevation in boiling point
B. Depression in freezing point
C. Osmotic pressure  ✓ Correct
D. Relative lowering of vapor pressure
Solution: Osmotic pressure is preferred for biomacromolecules because measurements are carried out at ambient temperature (preventing denaturation), and it yields measurable values even at millimolar concentrations.
Q3 — easy
The degree of dissociation ($\alpha$) of a weak monobasic acid in a $0.1\text{ M}$ solution having dissociation constant $K_a = 1.0 \times 10^{-5}$ is:
A. $0.001$
B. $0.01$  ✓ Correct
C. $0.1$
D. $0.005$
Solution: By Ostwald's dilution law: $\alpha = \sqrt{K_a/C} = \sqrt{10^{-5}/0.1} = \sqrt{10^{-4}} = 0.01$.
Q4 — easy
The enthalpy of vaporization of liquid water at $373\text{ K}$ is $40.66\text{ kJ/mol}$. The entropy change ($\Delta S_{\text{vap}}$) for the vaporization of 1 mole of water at $373\text{ K}$ is:
A. $109.0\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$  ✓ Correct
B. $92.4\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
C. $40.66\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
D. $1.09\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$
Solution: $\Delta S_{\text{vap}} = \Delta H_{\text{vap}}/T_b = 40660/373 \approx 109.0\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.
Q5 — easy
For a zero-order reaction $A \to \text{Products}$, which of the following expressions correctly relates the half-life period ($t_{1/2}$) to the initial reactant concentration $[A]_0$?
A. $t_{1/2} = \frac{0.693}{k}$
B. $t_{1/2} = \frac{[A]_0}{2k}$  ✓ Correct
C. $t_{1/2} = \frac{1}{k[A]_0}$
D. $t_{1/2} = \frac{[A]_0}{k}$
Solution: For zero-order kinetics: $[A] = [A]_0 - kt$. At $t_{1/2}$, $[A] = [A]_0/2$, so $t_{1/2} = [A]_0/(2k)$.
Q6 — easy
In the Contact process for the industrial manufacture of sulfuric acid, the heterogeneous catalyst used for the oxidation of $\text{SO}_2$ to $\text{SO}_3$ is:
A. Finely divided Fe
B. Platinized asbestos / $\text{V}_2\text{O}_5$  ✓ Correct
C. Ni powder
D. $\text{MnO}_2$
Solution: $\text{V}_2\text{O}_5$ or platinized asbestos operates at $720\text{ K}$ as the heterogeneous catalyst for the reversible oxidation of SO$_2$ to SO$_3$.
Q7 — easy
The correct decreasing order of acid strength among the oxoacids of chlorine is:
A. $\text{HClO}_4 > \text{HClO}_3 > \text{HClO}_2 > \text{HClO}$  ✓ Correct
B. $\text{HClO} > \text{HClO}_2 > \text{HClO}_3 > \text{HClO}_4$
C. $\text{HClO}_3 > \text{HClO}_4 > \text{HClO}_2 > \text{HClO}$
D. $\text{HClO}_4 > \text{HClO}_2 > \text{HClO}_3 > \text{HClO}$
Solution: As the oxidation state increases from +1 to +7, charge delocalization over oxygen atoms stabilizes the conjugate base, increasing acid strength: $\text{HClO}_4 > \text{HClO}_3 > \text{HClO}_2 > \text{HClO}$.
Q8 — easy
Which of the following properties is characteristic of interstitial compounds formed by transition metals?
A. Lower melting point than pure metals
B. Chemically more reactive than pure metals
C. Retain metallic electrical conductivity  ✓ Correct
D. Unusually soft and malleable
Solution: Interstitial compounds retain electrical and thermal conductivities close to the parent transition metal while acquiring higher hardness and higher melting points.
Q9 — easy
The hexadentate chelating ligand ethylenediaminetetraacetate ($\text{EDTA}^{4-}$) coordinates to a central metal ion through:
A. Two nitrogen and four oxygen atoms  ✓ Correct
B. Four nitrogen and two oxygen atoms
C. Six nitrogen atoms
D. Six oxygen atoms
Solution: $\text{EDTA}^{4-}$ coordinates via two tertiary amine nitrogens and four carboxylate oxygens, forming five stable five-membered chelate rings.
Q10 — easy
The coordination complex $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ exhibits geometrical isomerism. The cis-isomer acts pharmacologically as an effective:
A. Antibiotic
B. Anticancer drug (Cisplatin)  ✓ Correct
C. Analgesic
D. Antipyretic
Solution: Cisplatin, $\text{cis-}[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$, binds to purine bases of DNA in rapidly dividing cells, acting as a potent chemotherapy agent.
Q11 — easy
An alkyl chloride reacts with sodium iodide ($\text{NaI}$) in dry acetone to yield an alkyl iodide. This halogen exchange reaction is named:
A. Swarts reaction
B. Finkelstein reaction  ✓ Correct
C. Wurtz reaction
D. Frankland reaction
Solution: Finkelstein reaction: An alkyl chloride or bromide exchanges halogen with NaI in dry acetone (NaCl or NaBr precipitates, shifting equilibrium forward).
Q12 — easy
When an unknown alcohol is treated with Lucas reagent (anhydrous $\text{ZnCl}_2 + \text{conc. HCl}$) at room temperature, turbidity appears immediately within a few seconds. The alcohol is:
A. Ethanol
B. Propan-1-ol
C. Propan-2-ol
D. 2-Methylpropan-2-ol  ✓ Correct
Solution: 2-Methylpropan-2-ol is a tertiary alcohol. It forms a stable 3$^\circ$ carbocation rapidly with Lucas reagent, producing immediate cloudiness.
Q13 — easy
In the commercial cumene process for the manufacture of phenol, cumene (isopropylbenzene) is aerially oxidized to an intermediate which upon acidic hydrolysis yields phenol and:
A. Methanol
B. Acetaldehyde
C. Acetone (Propanone)  ✓ Correct
D. Acetic acid
Solution: Cumene auto-oxidation produces cumene hydroperoxide, which on treatment with dilute H$_2$SO$_4$ cleaves into phenol and propanone (acetone).
Q14 — easy
Benzene reacts with carbon monoxide and hydrogen chloride in the presence of anhydrous $\text{AlCl}_3$ and cuprous chloride ($\text{CuCl}$) to form benzaldehyde. This reaction is:
A. Etard reaction
B. Gattermann-Koch reaction  ✓ Correct
C. Stephen reaction
D. Rosenmund reduction
Solution: Gattermann-Koch reaction formylates benzene directly to benzaldehyde using CO + HCl under anhydrous AlCl$_3$/CuCl conditions.
Q15 — easy
Heating an anhydrous sodium salt of a carboxylic acid with soda-lime ($\text{NaOH} + \text{CaO}$, $3:1$) leads to the elimination of carbon dioxide, forming an alkane with:
A. One more carbon atom
B. One fewer carbon atom  ✓ Correct
C. The same number of carbon atoms
D. Twice the number of carbon atoms
Solution: Decarboxylation with soda-lime: $R\text{-COONa} + \text{NaOH} \to R\text{-H} + \text{Na}_2\text{CO}_3$, producing an alkane with one fewer carbon atom.
Q16 — easy
Which of the following substituted haloacetic acids exhibits the highest acidic strength?
A. $\text{FCH}_2\text{COOH}$  ✓ Correct
B. $\text{ClCH}_2\text{COOH}$
C. $\text{BrCH}_2\text{COOH}$
D. $\text{ICH}_2\text{COOH}$
Solution: Fluorine has the highest electronegativity, exerting the strongest $-I$ effect which stabilizes the carboxylate anion: $\text{FCH}_2\text{COOH}$ is the strongest acid.
Q17 — easy
Reaction of primary aliphatic amine with nitrous acid ($\text{NaNO}_2 + \text{HCl}$) at room temperature results in the quantitative evolution of which gas?
A. $\text{NH}_3$
B. $\text{NO}_2$
C. $\text{N}_2$  ✓ Correct
D. $\text{H}_2$
Solution: Primary aliphatic amines react with HNO$_2$ to form unstable diazonium salts that decompose immediately, releasing N$_2$ gas quantitatively.
Q18 — easy
Treatment of aniline with benzoyl chloride in the presence of aqueous sodium hydroxide to form benzanilide is an example of:
A. Carbylamine reaction
B. Schotten-Baumann reaction  ✓ Correct
C. Hofmann elimination
D. Gattermann reaction
Solution: Benzoylation of aromatic amines in the presence of aqueous alkali is the Schotten-Baumann reaction.
Q19 — easy
Hydrolysis of sucrose with dilute acid changes the optical sign of rotation from dextrorotatory to laevorotatory. The equimolar resulting mixture of D-(+)-glucose and D-(-)-fructose is termed:
A. Reducing sugar
B. Invert sugar  ✓ Correct
C. Malt sugar
D. Fruit sugar
Solution: Sucrose is dextrorotatory but on hydrolysis yields glucose (+52.5$^\circ$) and fructose (-92.4$^\circ$). Net rotation becomes laevorotatory, hence it is called invert sugar.
Q20 — easy
The dipolar ion structure of an amino acid containing both a protonated amino group and a deprotonated carboxylate group is known as a:
A. Carbocation
B. Zwitterion  ✓ Correct
C. Micelle
D. Radical ion
Solution: In neutral aqueous solution, amino acids exist as dipolar internal salts (zwitterions): $\text{H}_3\text{N}^+\text{-CH}(R)\text{-COO}^-$.
Q21 — easy
Buna-S is an elastomer synthesized by the emulsion copolymerization of 1,3-butadiene with:
A. Acrylonitrile
B. Styrene  ✓ Correct
C. Isoprene
D. Vinyl chloride
Solution: Buna-S (styrene-butadiene rubber, SBR) is synthesized from 1,3-butadiene and styrene in a 3:1 ratio.
Q22 — easy
Which of the following polymers is classified as a thermosetting plastic that undergoes irreversible cross-linking upon heating?
A. Polythene
B. Polyvinyl chloride (PVC)
C. Melamine-formaldehyde resin  ✓ Correct
D. Polystyrene
Solution: Melamine-formaldehyde resin is a thermosetting cross-linked polymer that hardens permanently upon molding and cannot be re-softened.
Q23 — easy
According to Green Chemistry principles, a chemical reaction displaying $100\%$ atom economy is exemplified by:
A. Addition of $\text{H}_2$ to ethylene to form ethane  ✓ Correct
B. Bromination of methane to form methyl bromide
C. Hydrolysis of ethyl acetate
D. Dehydration of ethanol to ethene
Solution: Catalytic hydrogenation $\text{CH}_2\text{=CH}_2 + \text{H}_2 \to \text{CH}_3\text{-CH}_3$ incorporates 100% of reactant atoms into the single product with no by-product waste.
Q24 — easy
Single-walled carbon nanotubes (SWCNTs) can be visualized as seamless hollow cylinders rolled up from a single planar sheet of:
A. Diamond
B. Graphene  ✓ Correct
C. Fullerene
D. Graphite oxide
Solution: SWCNTs are tubular structures formed by rolling up a single two-dimensional monolayer of $sp^2$ bonded graphene.
Q25 — easy
Reaction of an alkyl halide with silver fluoride ($\text{AgF}$) to yield an alkyl fluoride is known as:
A. Finkelstein reaction
B. Swarts reaction  ✓ Correct
C. Wurtz reaction
D. Ullmann reaction
Solution: Swarts reaction prepares alkyl fluorides by treating alkyl chlorides or bromides with heavy metal fluorides (AgF, Hg$_2$F$_2$, SbF$_3$).
Q26 — easy
Which functional group in carbohydrates is responsible for reducing Tollens' reagent and Fehling's solution?
A. Ester group
B. Free or hemiacetal aldehyde / alpha-hydroxy ketone group  ✓ Correct
C. Alcoholic ether linkage
D. Acetal glycosidic bond
Solution: Reducing sugars possess an open-chain aldehyde or an anomeric hemiacetal/hemiketal center capable of tautomerizing to an enediol that reduces Cu$^{2+}$ and Ag$^+$.
Q27 — easy
In the synthesis of ammonia: $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$, if $28\text{ g}$ of $\text{N}_2$ is reacted with $12\text{ g}$ of $\text{H}_2$, the mass of ammonia produced and the limiting reagent are:
A. $34\text{ g}$, $\text{N}_2$  ✓ Correct
B. $34\text{ g}$, $\text{H}_2$
C. $40\text{ g}$, $\text{N}_2$
D. $17\text{ g}$, $\text{H}_2$
Solution: Moles of N$_2$ = 28/28 = 1.0 mol; moles of H$_2$ = 12/2 = 6.0 mol. 1 mol N$_2$ requires 3 mol H$_2$. N$_2$ is limiting. NH$_3$ produced = 2 mol = 34 g.
Q28 — easy
Which of the following chemical processes represents a disproportionation redox reaction?
A. $2\text{H}_2\text{O}_2(aq) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g)$  ✓ Correct
B. $\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$
C. $\text{CH}_4 + 2\text{O}_2 \to \text{CO}_2 + 2\text{H}_2\text{O}$
D. $\text{NaCl} + \text{AgNO}_3 \to \text{AgCl} + \text{NaNO}_3$
Solution: In H$_2$O$_2$, oxygen has oxidation state $-1$. It is simultaneously reduced to $-2$ (in H$_2$O) and oxidized to $0$ (in O$_2$), which defines disproportionation.
Q29 — easy
Plaster of Paris, widely used in orthopedic casts, is chemically represented by the formula:
A. $\text{CaSO}_4\cdot 2\text{H}_2\text{O}$
B. $\text{CaSO}_4\cdot \frac{1}{2}\text{H}_2\text{O}$  ✓ Correct
C. $\text{CaSO}_4\cdot \text{H}_2\text{O}$
D. $\text{CaSO}_4$ (anhydrous)
Solution: Heating gypsum ($\text{CaSO}_4\cdot 2\text{H}_2\text{O}$) to $393\text{ K}$ produces the hemihydrate Plaster of Paris: $\text{CaSO}_4\cdot \frac{1}{2}\text{H}_2\text{O}$.
Q30 — easy
According to the Hardy-Schulze rule, which of the following coagulating ions is most effective for the coagulation of a negatively charged arsenic sulfide sol ($\text{As}_2\text{S}_3$)?
A. $\text{Na}^+$
B. $\text{Ba}^{2+}$
C. $\text{Al}^{3+}$  ✓ Correct
D. $\text{K}^+$
Solution: Hardy-Schulze rule: Coagulation power increases with the valency of the oppositely charged ion. For negatively charged As$_2$S$_3$ sol: Al$^{3+}$ > Ba$^{2+}$ > Na$^+$.