25 questions in 25 minutes on the NTA computer-based test interface, all from Some Basic Concepts of Chemistry. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Some Basic Concepts of Chemistry, of which 21 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
Revising first? Read the Some Basic Concepts of Chemistry chapter notes, then come back and take this test to check it stuck.
Loading your test…
Fetching the question paper. This takes a moment on a slow connection.
This test is part of the paid library
Unlock all 72 timed NEET CBT papers once, keep them forever. The 30 chapter tests stay free.
One payment. No subscription. Works on any device you log the key into.
Already bought it?
Paste the licence key from your payment email.
Duration: 180 minutes · 180 questions · 720 marks maximum
- The clock will be set at the server. The countdown timer at the top right of the screen will display the remaining time. When the timer reaches zero, the examination will end by itself.
- The Question Palette displayed on the right side of the screen will show the status of each question using one of the following symbols:
- Each correct answer carries +4 marks; each incorrect answer carries −1 mark. Un-attempted questions carry no marks.
- Navigate between sections using the section tabs, and between questions using SAVE & NEXT or the Question Palette.
- Your answers are saved in this browser, so an accidental refresh will not wipe a test in progress.
Submit the test?
Once you submit you cannot change answers. You will see your score and full solutions.
Your Result
Practice another mock test → · Continue practising on the ChapterNotes app →
Solutions & Revision
More Chemistry chapter tests
All NEET Chemistry CBT tests · The full NEET 2027 CBT mock library
NEET CBT mock test: common questions
How many questions are in this Some Basic Concepts of Chemistry mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 21 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Some Basic Concepts of Chemistry?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Chemistry
- Q1The number of significant figures for the three numbers 161 cm, 0.161 cm and 0.0161 cm are respectively:
- 3, 4 and 4
- 3, 3 and 4
- 3, 4 and 5
- 3, 3 and 3
Answer: (D) 3, 3 and 3
All non-zero digits are significant, and leading zeros (to the left of the first non-zero digit) are NOT significant. 161 has 3 s.f.; in 0.161 the leading 0 does not count, giving 3 s.f.; in 0.0161 the two leading zeros do not count, giving 3 s.f. So all three have 3 significant figures.
- Q2A temperature of 25 °C expressed in the SI base unit kelvin is:
- 248 K
- 273 K
- 300 K
- 298 K
Answer: (D) 298 K
K = °C + 273.15 ≈ °C + 273. So 25 °C = 25 + 273 = 298 K.
- Q3What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO₃(s) + 2HCl(aq) arrow CaCl₂(aq) + CO₂(g) + H₂O(l) [Calculate upto second place of decimal point]
- 1.32 g
- 9.50 g
- 3.65 g
- 1.25 g
Answer: (A) 1.32 g
B. Tests: stoichiometry followed by a purity correction applied in the right direction. Why B: moles of HCl = 0.050 × 0.5 = 0.025 mol. The equation needs 2 HCl per CaCO₃, so pure CaCO₃ required = 0.0125 mol × 100 g mol⁻¹ = 1.25 g. The sample is only 95% CaCO₃, so the mass to weigh out = 1.25/0.95 = 1.32 g. Why not A: 1.25 g is the pure CaCO₃ requirement, correct only if the sample were 100% pure; an impure sample always needs MORE [misconception A: “stopping before the purity correction”]. Why not C: 3.65 g is the molar mass of HCl (36.5) with the decimal shifted; no substitution of 0.025 mol into the 2:1 ratio with 100 g mol⁻¹ gives it. Why not D: 9.50 g merely echoes the 95 of the purity figure; it is 0.095 mol of pure CaCO₃, 7.6 times the 0.0125 mol actually needed. Remember: divide by the purity fraction, never multiply, because impurity means you must weigh out extra.
- Q4CaCO₃(s) + 2HCl(aq) arrow CaCl₂(aq) + CO₂(g) + H₂O(l). Consider the above reaction, what mass of CaCl₂ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃? (Given: molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol⁻¹, respectively)
- 10.545 g
- 2.636 g
- 3.908 g
- 5.272 g
Answer: (A) 10.545 g
C. Tests: spotting the limiting reagent when one reactant arrives as a solution and the other as a large solid mass. Why C: moles of HCl = 0.250 × 0.76 = 0.19 mol. Moles of CaCO₃ = 1000/100 = 10 mol, so HCl is limiting by a wide margin. The equation needs 2 HCl per CaCl₂, so CaCl₂ = 0.19/2 = 0.095 mol. Molar mass of CaCl₂ = 40 + 2(35.5) = 111 g mol⁻¹, so mass = 0.095 × 111 = 10.545 g. Why not A: 3.908 g is 0.0352 mol of CaCl₂; no consistent substitution of the 0.19 mol of HCl into the 2:1 ratio produces that number. Why not B: 2.636 g is 0.0238 mol, one quarter of the correct 0.095 mol, and nothing in the stoichiometry divides the HCl by eight. Why not D: 5.272 g is exactly half the answer, from applying the divide-by-2 of the 2:1 ratio a second time [misconception D: “halving twice because the coefficient is 2”]. Remember: 1000 g of a solid can still be the excess reagent; convert BOTH reactants to moles before deciding which one limits.
- Q5A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
- 1.4
- 3.0
- 2.8
- 4.4
Answer: (C) 2.8
C. Tests: combining dehydration products of two acids with selective absorption of an acidic gas. Why C: Formic acid HCOOH has molar mass 46, so 2.3/46 = 0.05 mol, and conc. H₂SO₄ dehydrates it to CO + H₂O, giving 0.05 mol CO. Oxalic acid H₂C₂O₄ has molar mass 90, so 4.5/90 = 0.05 mol, and it dehydrates to CO + CO₂ + H₂O, giving another 0.05 mol CO plus 0.05 mol CO₂. KOH pellets are basic and absorb only the acidic CO₂; CO is neutral and passes through. Remaining gas = 0.05 + 0.05 = 0.1 mol CO, mass = 0.1 x 28 = 2.8 g. Why not A: 1.4 g is 0.05 mol of CO, the CO from only one of the two acids; both acids release CO. Why not B: 3.0 g of CO would be 3.0/28 = 0.107 mol, which no combination of the two 0.05 mol batches produces; no consistent substitution gives this value. Why not D: 4.4 g is 0.1 mol of CO₂, which is what you get by assuming both acids give only CO₂ and that KOH removes the CO [misconception D: “KOH absorbs whatever gas is left”]; KOH absorbs CO₂, not CO. Remember: conc. H₂SO₄ takes water out, formic acid leaves CO while oxalic acid leaves CO + CO₂, and KOH keeps only the CO₂.
- Q6An element X has the isotopic composition ²⁰⁰X : 90%, ¹⁹⁹X : 8.0% and ²⁰²X : 2.0%. The weighted average atomic mass of naturally occurring X is closest to:
- 199 u
- 202 u
- 201 u
- 200 u
Answer: (D) 200 u
Average = 200×0.90 + 199×0.08 + 202×0.02 = 180.0 + 15.92 + 4.04 = 199.96 ≈ 200 u.
- Q7For ionic compounds like NaCl, the correct term for the sum of atomic masses in the formula unit is:
- formula mass
- atomic mass
- equivalent mass
- molecular mass
Answer: (A) formula mass
Ionic compounds do not exist as discrete molecules; they form a continuous lattice. Hence we use ‘formula mass’ (formula unit mass), not molecular mass, for compounds such as NaCl (58.5 u).
- Q8The number of water molecules is maximum in
- 1.8 gram of water
- 18 molecules of water
- 18 gram of water
- 18 moles of water
Answer: (D) 18 moles of water
C. Tests: comparing quantities stated in grams, moles and raw molecule counts on one scale. Why C: Convert each to moles. 1.8 g = 1.8/18 = 0.1 mol. 18 g = 18/18 = 1 mol. 18 moles is 18 mol outright. 18 molecules = 18/(6.022 x 10²³) = 3 x 10⁻²³ mol. The largest is 18 mol, which is 18 x 6.022 x 10²³ = 1.08 x 10²⁵ molecules. Why not A: 1.8 g is 0.1 mol, 180 times fewer molecules than 18 mol. Why not B: 18 g is exactly 1 mol, still 18 times fewer than 18 mol [misconception B: “18 grams and 18 moles are the same thing for water”]. Why not D: 18 molecules is literally eighteen particles, vanishingly small next to a mole. Remember: the unit attached to the number decides everything, so convert grams and molecule counts to moles before comparing.
- Q9The number of moles of oxygen in 1 L of air containing 21% oxygen by volume, under standard conditions, is:
- 0.21 mole
- 0.186 mole
- 0.0093 mole
- 2.10 moles
Answer: (C) 0.0093 mole
Volume of O₂ in 1 L of air = (21/100) × 1000 mL = 210 mL. At STP, 22400 mL of any gas is 1 mole, so moles of O₂ = 210 / 22400 = 0.0093 mol.
- Q10What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution? [Given: Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol⁻¹ respectively.]
- 0.867
- 0.967
- 0.825
- 0.032
Answer: (B) 0.967
D. Tests: mass percent to moles to mole fraction, asked for the solvent. Why D: take 100 g of solution, so 10 g of urea and 90 g of water. Urea NH₂CONH₂ has molar mass 60, giving 10/60 = 0.1667 mol. Water gives 90/18 = 5 mol. Mole fraction of water = 5/(5 + 0.1667) = 5/5.1667 = 0.967. Why not B: 0.032 is 0.1667/5.1667, the mole fraction of urea, the solute, not of the water asked for [misconception: “the mole fraction always refers to the solute”]. Why not A: no consistent substitution of the given molar masses gives 0.825; the only two mole fractions this data supports are 0.967 and 0.032. Why not C: 0.867 is also unreachable, since matching it would need 0.77 mol of urea in 90 g of water, that is a urea molar mass of 13 g mol⁻¹. Remember: mole fractions add to 1, so compute the easier component and subtract.
- Q11The percentage weight of Zn in white vitriol [ZnSO₄·7H₂O] is approximately equal to (atomic masses: Zn = 65, S = 32, O = 16, H = 1):
- 23.65%
- 33.65%
- 32.56%
- 22.65%
Answer: (D) 22.65%
Molar mass of ZnSO₄·7H₂O = 65 + 32 + (4×16) + 7×18 = 65 + 32 + 64 + 126 = 287. %Zn = (65/287) × 100 = 22.65%.
- Q12On combustion 0.210 g of an organic compound containing C, H and O gave 0.127 g H₂O and 0.307 g CO₂. The percentages of hydrogen and oxygen in the given organic compound respectively are:
- 53.41, 39.6
- 6.72, 39.87
- 6.72, 53.41
- 7.55, 43.85
Answer: (C) 6.72, 53.41
B. Tests: combustion analysis, turning CO₂ and H₂O masses into element percentages with oxygen taken by difference. Why B: mass of H = (2/18) × 0.127 = 0.01411 g, so %H = (0.01411/0.210) × 100 = 6.72%. Mass of C = (12/44) × 0.307 = 0.08373 g, so %C = (0.08373/0.210) × 100 = 39.87%. Oxygen is never measured directly, so %O = 100 – 6.72 – 39.87 = 53.41%. The pair asked for is (6.72, 53.41). Why not A: 7.55% hydrogen needs 0.01586 g of H, which would have given 0.1427 g of water, not the 0.127 g collected; its partner 43.85 is neither the 39.87% carbon nor the 53.41% oxygen, so neither figure of the pair follows from the data. Why not C: 6.72 is right for hydrogen, but 39.87 is the percentage of CARBON, not oxygen [misconception C: “quoting %C in the oxygen slot, forgetting O is the by-difference element”]. Why not D: these are the right numbers in the wrong order, 53.41 (which is oxygen) placed as hydrogen and a rounded 39.6 (which is carbon) placed as oxygen. Remember: in combustion analysis only C and H come from the products; oxygen is always 100 minus everything else.
- Q13An organic compound contains 78% (by weight) carbon and the remaining percentage of hydrogen. The empirical formula of this compound is (atomic masses: C = 12, H = 1):
- CH₂
- CH
- CH₄
- CH₃
Answer: (D) CH₃
%H = 100 − 78 = 22. Relative moles: C = 78/12 = 6.5; H = 22/1 = 22. Divide by the smaller (6.5): C = 1, H = 22/6.5 ≈ 3.4 ≈ 3. Empirical formula = CH₃.
- Q14Dalton’s Atomic theory could not explain which of the following?
- Law of conservation of mass
- Law of gaseous volume
- Law of constant proportion
- Law of multiple proportion
Answer: (B) Law of gaseous volume
B. Tests: knowing which combination law sits outside the reach of Dalton’s atomic theory. Why B: Dalton treated the atom as the indivisible unit and had no concept of diatomic molecules, so he could not account for Gay-Lussac’s law of gaseous volumes, where gases react in simple whole-number volume ratios. Explaining it needed Avogadro’s molecular hypothesis. Why not A: the law of multiple proportions falls straight out of Dalton’s theory, atoms combining in different small whole-number ratios, and Dalton used it as evidence for it. Why not C: conservation of mass follows from atoms being indestructible and merely rearranged during a reaction. Why not D: the law of constant proportion follows from a compound containing a fixed number of each kind of atom [misconception: “Dalton’s theory failed for all the combination laws”]. Remember: Dalton handles the mass laws, Avogadro is needed for the volume law.
- Q15What is the mole fraction of the solute in a 1.00 m aqueous solution?
- 0.0354
- 1.770
- 0.177
- 0.0177
Answer: (D) 0.0177
C. Tests: converting molality into mole fraction through the moles of solvent. Why C: 1.00 molal means 1 mol of solute in 1000 g of water. Moles of water = 1000/18 = 55.5 mol. Mole fraction of solute = 1/(1 + 55.5) = 1/56.5 = 0.0177. Why not A: 1.770 is 100 times the correct value, a two-place decimal slip, and it is impossible anyway because a mole fraction can never exceed 1. Why not D: 0.177 is exactly 10 times the correct value, a one-place decimal slip. Why not B: 0.0354 is exactly twice the correct value, since 2/56.5 = 0.0354, so it is the answer for a 2 molal solution rather than the 1.00 molal one asked about; keeping 1 mol of solute and solving 1/(1+n) = 0.0354 would need n = 27.25 mol of water, about 490 g, not the 1000 g that molality fixes [misconception B: “molality and mole fraction both count the solute twice somewhere”]. Remember: 1 kg of water is 55.5 mol, so a 1 molal solution always has a solute mole fraction near 1/56.5.
- Q16What fraction of Fe exists as Fe(III) in Fe_(0.96)O? (Consider Fe_(0.96)O to be made up of Fe(II) and Fe(III) only)
- (1)/(16)
- 0.08
- (1)/(20)
- (1)/(12)
Answer: (D) (1)/(12)
B. Tests: charge balance in a non-stoichiometric oxide, and reading “fraction of Fe” as a fraction of the iron actually present. Why B: let x mol of Fe(III) and y mol of Fe(II) sit in one formula unit, so x + y = 0.96. One O²⁻ demands +2 of positive charge, so 3x + 2y = 2. Substituting y = 0.96 – x gives 3x + 1.92 – 2x = 2, hence x = 0.08 mol of Fe(III). The fraction of the iron that is Fe(III) = 0.08/0.96 = 1/12. Why not A: 1/20 = 0.05 means 0.048 mol of Fe(III) and 0.912 mol of Fe(II), a total charge of 3(0.048) + 2(0.912) = 1.968, so the oxide would not be neutral. Why not C: 0.08 is the AMOUNT of Fe(III) per formula unit, not its fraction of the iron; you still have to divide it by the 0.96 total [misconception C: “reporting the mole amount as the fraction”]. Why not D: 1/16 = 0.0625 means 0.06 mol of Fe(III) and 0.90 mol of Fe(II), a total charge of 0.18 + 1.80 = 1.98, again short of the +2 required. Remember: set total positive charge equal to total negative charge, then divide the Fe(III) amount by the TOTAL Fe, not by 1.
- Q17Concentrated nitric acid is labelled as 75% by mass. The volume in mL of the solution which contains 30 g of nitric acid is ______________. (Given: density of nitric acid solution is 1.25 g/mL)
- 55
- 45
- 32
- 40
Answer: (C) 32
A. Tests: mass percent to solution mass, then density to convert that mass into a volume. Why A: 75% by mass means 75 g of HNO₃ in every 100 g of solution, so 30 g of acid sits inside 30/0.75 = 40 g of solution. Volume = mass/density = 40/1.25 = 32 mL. Why not B: 40 is the MASS of the solution in grams; it only becomes a volume after dividing by the density 1.25 g/mL [misconception B: “treating grams of solution as millilitres”]. Why not C: 55 mL of this solution weighs 55 × 1.25 = 68.75 g and would carry 0.75 × 68.75 = 51.6 g of acid, well above the 30 g specified. Why not D: 45 mL weighs 56.25 g and carries 42.2 g of acid; no combination of 75% and 1.25 g/mL puts 30 g of acid into 45 mL. Remember: percent by mass hands you the solution’s MASS, and only the density turns that mass into a volume.
- Q18The volume occupied by one molecule of water (density = 1 g cm⁻³) is approximately:
- 5.5×10⁻²³ cm³
- 3.0×10⁻²³ cm³
- 6.023×10⁻²³ cm³
- 9.0×10⁻²³ cm³
Answer: (B) 3.0×10⁻²³ cm³
1 mole (18 g) of water contains 6.023×10²³ molecules, so the mass of one molecule = 18 / 6.023×10²³ g. With density 1 g cm⁻³, volume = mass = 18 / 6.023×10²³ ≈ 3.0×10⁻²³ cm³.
- Q19Which pair of compounds has the same percentage composition by mass and the same empirical formula?
- CO and CO₂
- Glucose (C₆H₁₂O₆) and acetic acid (CH₃COOH)
- Methane and ethane
- Water and hydrogen peroxide
Answer: (B) Glucose (C₆H₁₂O₆) and acetic acid (CH₃COOH)
Glucose C₆H₁₂O₆ and acetic acid C₂H₄O₂ both reduce to the empirical formula CH₂O, so they share the same empirical formula and identical percentage composition (40% C, 6.7% H, 53.3% O).
- Q20The mass of iron converted into Fe₃O₄ by the action of 18 g of steam is: (Given: Molar mass of H, O and Fe are 1, 16 and 56 g mol⁻¹ respectively) Assume iron is present in excess:
- 21 g
- 42 g
- 4.2 g
- 2.1 g
Answer: (B) 42 g
D. Tests: writing the steam-on-iron equation and using its mole ratio with steam as the limiting reagent. Why D: the reaction is 3Fe + 4H₂O arrow Fe₃O₄ + 4H₂. Steam = 18/18 = 1 mol. Iron consumed = (3/4) x 1 = 0.75 mol, so mass = 0.75 x 56 = 42 g. Why not C: 21 g is 0.375 mol of Fe, which follows if the steam is counted as 0.5 mol, that is a water molar mass of 36 instead of 18 [misconception: “water weighs 36 because hydrogen is diatomic”]. Why not B: 4.2 g is the correct 42 g with the decimal moved one place, and no ratio in the balanced equation turns 1 mol of steam into 0.075 mol of Fe. Why not A: 2.1 g is 0.0375 mol of Fe, one twentieth of the required 0.75 mol, and nothing in the 3 : 4 ratio produces a twentieth. Remember: magnetite needs 3 Fe to 4 H₂O, so moles of iron = 0.75 x moles of steam.
- Q21The number of molecules and moles in 2.8375 litres of O₂ at STP are respectively
- 1.505 × 10²³ and 0.250 mol
- 7.527 × 10²² and 0.125 mol
- 7.527 × 10²² and 0.250 mol
- 7.527 × 10²³ and 0.125 mol
Answer: (B) 7.527 × 10²² and 0.125 mol
D. Tests: converting a gas volume at STP to moles with the current molar volume (22.7 L), then moles to molecules via N_A, without sliding from molecules to atoms. Why D: at STP (273.15 K, 1 bar) one mole of an ideal gas occupies 22.7 L, so n = (2.8375)/(22.7) = 0.125 mol. Molecules = 0.125 × 6.022 × 10²³ = 7.5275 × 10²² ≈ 7.527 × 10²². So 7.527 × 10²² molecules and 0.125 mol. Why not A: both numbers here are the atom count, not the molecule count. O₂ is diatomic, so 0.125 mol of O₂ contains 2 × 0.125 = 0.250 mol of O atoms, and 0.250 × 6.022 × 10²³ = 1.505 × 10²³ atoms. The question asks for molecules and moles of the gas, so doubling for the two atoms per molecule is exactly the wrong move [misconception A: “counting atoms instead of molecules for a diatomic gas”]. Why not B: the two halves come from different routes. 7.527 × 10²² is the correct molecule count, which belongs to 0.125 mol, while 0.250 mol is the atom figure; check them against each other and 0.250 × 6.022 × 10²³ = 1.505 × 10²³, not 7.527 × 10²² [misconception B: “reporting a molecule count alongside a mole value that would not produce it”]. Why not C: the mole value 0.125 is right but the exponent is one place too high. 0.125 × 6.022 × 10²³ = 7.527 × 10²², whereas 7.527 × 10²³ would correspond to (7.527 × 10²³)/(6.022 × 10²³) = 1.25 mol, ten times the gas actually present [misconception C: “keeping the 10²³ of N_A unchanged when the coefficient drops below 1″]. Remember: at STP use 22.7 L per mole, multiply moles by 6.022 × 10²³ for molecules, and double only if the question asks for atoms.
- Q22Find the number of molecules present in 70 g dinitrogen.
- 1.5055 × 10²³
- 1.5055 × 10²⁴
- 3.011 × 10²³
- 3.011 × 10²⁴
Answer: (B) 1.5055 × 10²⁴
A. Tests: mass to moles to molecule count for a diatomic gas. Why A: dinitrogen is N₂, molar mass 28 g mol⁻¹. Moles = 70/28 = 2.5. Molecules = 2.5 × 6.022 × 10²³ = 1.5055 × 10²⁴. Why not B: 1.5055 × 10²³ carries the right coefficient with the power ten times too small; it corresponds to 0.25 mol, which would be 7 g of N₂. Why not C: 3.011 × 10²⁴ is 5 mol, exactly what 70/14 gives if the atomic mass 14 is used instead of the molecular mass 28 [misconception C: “treating dinitrogen as single N atoms”]. Why not D: 3.011 × 10²³ is 0.5 mol, that is 14 g of N₂; no substitution of 70 g with 28 g mol⁻¹ lands there. Remember: dinitrogen means N₂ at 28 g mol⁻¹, and the question asks for molecules, not atoms.
- Q23What volume of hydrogen gas at STP would be liberated by action of 50 mL of H₂SO₄ of 50% purity (density = 1.3 g mL⁻¹) on 20 g of zinc? Given: Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol⁻¹ respectively.
- 7.428 L
- 8.375 L
- 6.892 L
- 5.824 L
Answer: (C) 6.892 L
C. Tests: density and purity to moles of acid, then a limiting-reagent comparison against zinc. Why C: mass of acid solution = 50 x 1.3 = 65 g, and 50% purity leaves 32.5 g of H₂SO₄, that is 32.5/98 = 0.332 mol. Zinc supplies 20/65 = 0.308 mol. In Zn + H₂SO₄ arrow ZnSO₄ + H₂ the ratio is 1 : 1, so zinc runs out first and H₂ = 0.308 mol. At STP that is 0.308 x 22.4 = 6.892 L. Why not B: 7.428 L is 0.332 x 22.4, the volume you get by letting all the acid react and ignoring that zinc is the smaller supply [misconception: “the reagent given in the bigger amount controls the yield”]. Why not A: no consistent substitution of the given data produces 5.824 L; it corresponds to 0.26 mol of H₂, which neither reactant supplies. Why not D: no consistent substitution produces 8.375 L either; it is 0.374 mol of H₂, more than even the acid-limited 0.332 mol, so no reading of the data can justify it. Remember: density and purity first, moles second, and the smaller mole supply always wins.
- Q24A compound contains 24 g carbon, 4 g hydrogen and 32 g oxygen. Its empirical formula is (C=12, H=1, O=16):
- CHO
- CH₄O
- CH₂O
- C₂H₄O
Answer: (C) CH₂O
Moles: C = 24/12 = 2, H = 4/1 = 4, O = 32/16 = 2. Divide each by the smallest value (2): C = 1, H = 2, O = 1. The ratio C:H:O = 1:2:1, so the empirical formula is CH₂O.
- Q25A + 2B arrow AB₂ 36.0 g of ‘A’ (Molar mass: 60 g mol⁻¹) and 56.0 g of ‘B’ (Molar mass: 80 g mol⁻¹) are allowed to react. Which of the following statements are correct? A. ‘A’ is the limiting reagent. B. 77.0 g of AB₂ is formed. C. Molar mass of AB₂ is 140 g mol⁻¹. D. 15.0 g of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below:
- B and D Only
- A and B Only
- A and C Only
- C and D Only
Answer: (A) B and D Only
D. Tests: limiting reagent under a 1 : 2 stoichiometry, then product mass and leftover reactant. Why D: moles of A = 36.0/60 = 0.6 and moles of B = 56.0/80 = 0.7. The equation needs 2 B per A, so 0.6 mol of A would demand 1.2 mol of B while only 0.7 mol is present, making B the limiting reagent. That 0.7 mol of B consumes 0.35 mol of A and gives 0.35 mol of AB₂. Molar mass of AB₂ = 60 + 2 x 80 = 220, so mass = 0.35 x 220 = 77.0 g, which is statement B. A left over = 0.6 – 0.35 = 0.25 mol = 15.0 g, which is statement D. Why not A: statement A is false because A is in excess, not limiting, and statement C is false because 60 + 2 x 80 = 220 g mol⁻¹, not 140 [misconception: “the reactant with fewer moles is automatically limiting”]. Why not B: its statement B (77.0 g) is right, but it carries the false statement A along with it. Why not C: its statement D (15.0 g of A left) is right, but statement C adds only one B to A instead of two when computing the molar mass. Remember: divide each reactant’s moles by its own coefficient, and the smaller quotient is the limiting reagent.