JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 30

Multiple Choice Questions (MCQs) for JEE Chemistry: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: Under the same conditions of temperature and pressure, two gases contain the same number of molecules. They must
(a) be noble gases
(b) have a volume of 22.4 dm3 each
(c) have an equal number of atoms
(d) have equal volumes
Show Answer & Explanation

Answer: (d) have equal volumes

Explanation:
1. By Avogadro's hypothesis, equal volumes of gases at the same T and P contain equal numbers of molecules, and the converse is also true.
2. The volume of 22.4 dm3 applies only at STP, and gases need not be monoatomic, so the atom count may differ.

Teacher's Note:
1. Avogadro's law counts molecules, not atoms.
2. Molar volume (22.4 L) is quoted only at 273 K and 1 atm.

Question 2: The empirical formula of a compound is CH2O and its molecular mass is 120 u. The molecular formula is
(a) C2H4O2
(b) C4H8O4
(c) C3H6O3
(d) all of these
Show Answer & Explanation

Answer: (b) C4H8O4

Explanation:
1. Empirical formula mass of CH2O = 12 + 2 + 16 = 30 u.
2. n = 120 / 30 = 4.
3. Molecular formula = C4H8O4. The others have masses 60 u and 90 u.

Teacher's Note:
1. The molecular formula is n times the empirical formula, n = molecular mass / empirical mass.
2. Different multiples give different molecular masses, so only one fits.

Question 3: A compound contains 20% Ca and 80% Br by mass and its molecular mass is 200 u (Ca = 40, Br = 80). Its molecular formula is
(a) Ca2Br
(b) CaBr
(c) Ca1/2Br
(d) CaBr2
Show Answer & Explanation

Answer: (d) CaBr2

Explanation:
1. Moles in 100 g: Ca = 20/40 = 0.5, Br = 80/80 = 1.
2. Ratio Ca : Br = 1 : 2, so the empirical formula is CaBr2 with mass 40 + 160 = 200 u.
3. This equals the given molecular mass, so the molecular formula is CaBr2.

Teacher's Note:
1. Divide the percentage by the atomic mass, then divide by the smallest value.
2. Compare the empirical mass with the molecular mass.

Question 4: A compound contains 8% sulphur by mass. The least possible molecular mass of the compound is
(a) 400 u
(b) 155 u
(c) 355 u
(d) 200 u
Show Answer & Explanation

Answer: (a) 400 u

Explanation:
1. A molecule must contain at least one S atom (32 u).
2. If 8% of the mass is 32 u, the molecular mass = 32 / 0.08 = 400 u.

Teacher's Note:
1. Minimum molecular mass corresponds to the smallest whole number of atoms of the element.
2. Mass of that atom divided by its mass fraction gives the answer.

Question 5: How many moles of potassium chlorate must be heated to produce 11.2 L of oxygen at STP? KClO3 → KCl + 3/2 O2
(a) 1/4 mol
(b) 1/2 mol
(c) 2/3 mol
(d) 1/3 mol
Show Answer & Explanation

Answer: (d) 1/3 mol

Explanation:
1. Moles of O2 = 11.2 / 22.4 = 0.5.
2. 1 mol KClO3 gives 3/2 mol O2, so moles of KClO3 = 0.5 × 2/3 = 1/3 mol.

Teacher's Note:
1. Use the mole ratio from the balanced equation.
2. Convert the gas volume to moles first.

Question 6: For the reaction 2P + Q → R, 8 mol of P and excess of Q will produce
(a) 8 mol of R
(b) 4 mol of R
(c) 5 mol of R
(d) 13 mol of R
Show Answer & Explanation

Answer: (b) 4 mol of R

Explanation:
1. 2 mol P gives 1 mol R.
2. 8 mol P gives 8 / 2 = 4 mol R.

Teacher's Note:
1. Q is in excess, so P decides the yield.
2. Divide by the coefficient of the limiting reactant.

Question 7: If 1.5 mol of oxygen combines with aluminium to form Al2O3, the mass of aluminium used is
(a) 54 g
(b) 27 g
(c) 40.5 g
(d) 81 g
Show Answer & Explanation

Answer: (a) 54 g

Explanation:
1. 4Al + 3O2 → 2Al2O3.
2. Moles of Al = 1.5 × 4/3 = 2 mol.
3. Mass = 2 × 27 = 54 g.

Teacher's Note:
1. Balance the equation before using the mole ratio.
2. Convert moles to mass with the atomic mass.

Question 8: When 100 g of ethylene polymerises completely to polyethene, n(C2H4) → (CH2CH2)n, the mass of polyethene formed is
(a) (n/2) g
(b) 100 g
(c) (100/n) g
(d) 100n g
Show Answer & Explanation

Answer: (b) 100 g

Explanation:
1. Polymerisation only joins molecules; no atom is lost.
2. By conservation of mass, 100 g of monomer gives 100 g of polymer.

Teacher's Note:
1. Addition polymerisation has no by-product.
2. Mass is conserved in every chemical reaction.

Question 9: 12 g of an alkaline earth metal gives 14.8 g of its nitride. The atomic mass of the metal is
(a) 40
(b) 20
(c) 12
(d) 14.8
Show Answer & Explanation

Answer: (a) 40

Explanation:
1. Alkaline earth metals form nitrides of formula M3N2.
2. Mass of nitrogen combined = 14.8 - 12 = 2.8 g = 0.2 mol N atoms.
3. Moles of metal = 0.2 × 3/2 = 0.3 mol, so atomic mass = 12 / 0.3 = 40.

Teacher's Note:
1. Use the formula M3N2 for a group 2 nitride.
2. Find moles of the known element, then of the unknown.

Question 10: 0.5 mol of H2SO4 is mixed with 0.2 mol of Ca(OH)2. The maximum number of moles of CaSO4 formed is
(a) 0.4
(b) 1.5
(c) 0.2
(d) 0.5
Show Answer & Explanation

Answer: (c) 0.2

Explanation:
1. H2SO4 + Ca(OH)2 → CaSO4 + 2H2O, ratio 1 : 1.
2. Ca(OH)2 (0.2 mol) is the limiting reagent.
3. Moles of CaSO4 = 0.2.

Teacher's Note:
1. Compare mole/coefficient ratios to find the limiting reagent.
2. Product moles follow the limiting reagent.

Question 11: Equal masses of X (atomic mass 36) and Y (atomic mass 24) react to form X2Y3. Then
(a) X is the limiting reagent
(b) Y is the limiting reagent
(c) no reactant is left over and the mass of X2Y3 formed is double the mass of X taken
(d) none of these
Show Answer & Explanation

Answer: (c) no reactant is left over and the mass of X2Y3 formed is double the mass of X taken

Explanation:
1. Take 72 g of each: X = 2 mol, Y = 3 mol.
2. X2Y3 needs X : Y = 2 : 3, exactly the available ratio, so both are consumed.
3. Mass of X2Y3 = 72 + 72 = 144 g = 2 x mass of X.

Teacher's Note:
1. Convert equal masses to moles before comparing.
2. When the ratio matches exactly, neither reagent is in excess.

Question 12: Nickel is extracted by Mond's process: Ni + 4CO → Ni(CO)4. If the CO is made from 6 g of carbon and 44 g of CO2 (C + CO2 → 2CO), the mass of nickel needed (Ni = 59 u) is
(a) 14.75 g
(b) 58 g
(c) 29 g
(d) 28 g
Show Answer & Explanation

Answer: (a) 14.75 g

Explanation:
1. Moles of C = 6/12 = 0.5; moles of CO2 = 44/44 = 1.
2. Carbon is limiting, so CO formed = 2 × 0.5 = 1 mol.
3. Moles of Ni = 1/4 = 0.25 mol, mass = 0.25 × 59 = 14.75 g.

Teacher's Note:
1. Chain two reactions through the intermediate CO.
2. Find the limiting reagent of the first step.

Question 13: What mass of CaCO3 must be decomposed to give enough CO2 to convert 21.2 kg of Na2CO3 completely into NaHCO3? (CaCO3 → CaO + CO2; Na2CO3 + CO2 + H2O → 2NaHCO3; Na = 23, Ca = 40)
(a) 30 kg
(b) 20 kg
(c) 120 kg
(d) 100 kg
Show Answer & Explanation

Answer: (b) 20 kg

Explanation:
1. Moles of Na2CO3 = 21 200 / 106 = 200 mol.
2. CO2 needed = 200 mol, so CaCO3 needed = 200 mol.
3. Mass = 200 × 100 g = 20 kg.

Teacher's Note:
1. Use the 1 : 1 : 1 ratios through CO2.
2. Keep units consistent (g to kg).

Question 14: Iodine is obtained from brine by NaI + AgNO3 → AgI + NaNO3; 2AgI + Fe → FeI2 + 2Ag; 2FeI2 + 3Cl2 → 2FeCl3 + 2I2. What mass of AgNO3 is needed in the first step for every 254 kg of I2 produced in the third step? (Ag = 108, I = 127, N = 14)
(a) 68 kg
(b) 85 kg
(c) 380 kg
(d) 340 kg
Show Answer & Explanation

Answer: (d) 340 kg

Explanation:
1. 254 kg I2 = 1000 mol I2, which needs 1000 mol FeI2.
2. 1000 mol FeI2 needs 2000 mol AgI, hence 2000 mol AgNO3.
3. Mass = 2000 × 170 g = 340 kg.

Teacher's Note:
1. Link the steps by the common intermediate, I atoms.
2. Each I2 contains two I atoms, each from one AgNO3.

Question 15: 25.4 g of iodine and 14.2 g of chlorine react completely to give a mixture of ICl and ICl3. The moles of ICl and ICl3 formed are
(a) 0.1 mol, 0.2 mol
(b) 0.5 mol, 0.5 mol
(c) 0.1 mol, 0.1 mol
(d) 0.2 mol, 0.2 mol
Show Answer & Explanation

Answer: (c) 0.1 mol, 0.1 mol

Explanation:
1. I atoms = 2 × 25.4/254 = 0.2 mol; Cl atoms = 2 × 14.2/71 = 0.4 mol.
2. Let ICl = x and ICl3 = y: x + y = 0.2 and x + 3y = 0.4.
3. Solving gives y = 0.1 and x = 0.1.

Teacher's Note:
1. Write one atom-balance equation per element.
2. Solve the two simultaneous equations.

Question 16: What masses of P4O6 and P4O10 are produced by burning 31 g of P4 in 32 g of oxygen, leaving neither P4 nor O2?
(a) 27.5 g, 35.5 g
(b) 17.5 g, 190.5 g
(c) 2.75 g, 219.5 g
(d) 55 g, 71 g
Show Answer & Explanation

Answer: (a) 27.5 g, 35.5 g

Explanation:
1. Moles of P4 = 31/124 = 0.25; moles of O2 = 1.
2. Let P4O6 = a and P4O10 = b: a + b = 0.25 and 3a + 5b = 1.
3. Solving gives a = b = 0.125 mol.
4. Masses: P4O6 = 0.125 × 220 = 27.5 g; P4O10 = 0.125 × 284 = 35.5 g.

Teacher's Note:
1. Balance P atoms and O2 molecules separately.
2. P4O6 needs 3 O2 and P4O10 needs 5 O2 per P4.

Question 17: An aqueous solution of H2PO4- is treated with Mg2+ and NH4+ to precipitate Mg(NH4)PO4·6H2O, which is heated to Mg2P2O7. A solution gave 1.054 g of Mg2P2O7 (molar mass 222.5 g). The mass of NaH2PO4 (120 g mol-1) originally present was
(a) 1.62 g
(b) 1.14 g
(c) 2.34 g
(d) 1.33 g
Show Answer & Explanation

Answer: (b) 1.14 g

Explanation:
1. Moles of Mg2P2O7 = 1.054 / 222.5 = 4.737 × 10-3 mol.
2. Each Mg2P2O7 contains 2 P atoms, so moles of P = 9.47 × 10-3 mol = moles of NaH2PO4.
3. Mass = 9.47 × 10-3 x 120 = 1.14 g.

Teacher's Note:
1. Count P atoms: 2 per pyrophosphate.
2. Convert moles back to mass with the original salt's molar mass.

Question 18: A 10 g mixture of CaCl2 and NaCl is treated to precipitate all calcium as CaCO3, which is heated to CaO. The final mass of CaO is 1.62 g. The percentage by mass of CaCl2 in the mixture is
(a) 11.0%
(b) 32.1%
(c) 21.8%
(d) 16.2%
Show Answer & Explanation

Answer: (b) 32.1%

Explanation:
1. Moles of CaO = 1.62 / 56 = 0.0289 mol = moles of CaCl2.
2. Mass of CaCl2 = 0.0289 × 111 = 3.21 g.
3. Percentage = 3.21 / 10 × 100 = 32.1%.

Teacher's Note:
1. Ca atoms are conserved from CaCl2 to CaO.
2. Use molar mass 111 g for CaCl2.

Question 19: The mass of 70% pure H2SO4 required for the neutralisation of 1 mol of NaOH is
(a) 70 g
(b) 34.3 g
(c) 98 g
(d) 49 g
Show Answer & Explanation

Answer: (a) 70 g

Explanation:
1. H2SO4 + 2NaOH → Na2SO4 + 2H2O, so 1 mol NaOH needs 0.5 mol H2SO4 = 49 g pure acid.
2. Impure sample mass = 49 / 0.70 = 70 g.

Teacher's Note:
1. Pure mass needed divided by the purity fraction gives sample mass.
2. One mole of H2SO4 neutralises two moles of NaOH.

Question 20: What volume of a 0.8 M solution contains 100 millimoles of the solute?
(a) 100 mL
(b) 62.5 mL
(c) 500 mL
(d) 125 mL
Show Answer & Explanation

Answer: (d) 125 mL

Explanation:
1. V = moles / molarity = 0.1 mol / 0.8 mol L-1 = 0.125 L = 125 mL.

Teacher's Note:
1. Molarity x volume (L) = moles; mmol / M gives mL directly.
2. 100 / 0.8 = 125.

Question 21: A solution of FeCl3 is M/30. The molarity of Cl- ions in it is
(a) M/90
(b) M/10
(c) M/30
(d) M/5
Show Answer & Explanation

Answer: (b) M/10

Explanation:
1. FeCl3 dissociates completely: one formula unit gives three Cl- ions.
2. [Cl-] = 3 × M/30 = M/10.

Teacher's Note:
1. Multiply the salt concentration by the number of ions per formula unit.
2. Assume complete dissociation.

Question 22: A solution contains equal moles of H2O and NaCl. The molality of the NaCl is
(a) 55.5
(b) 1.00
(c) 0.55
(d) 0.18
Show Answer & Explanation

Answer: (a) 55.5

Explanation:
1. Take 1 mol NaCl and 1 mol water (18 g = 0.018 kg).
2. Molality = 1 / 0.018 = 55.5 mol kg-1.

Teacher's Note:
1. Molality uses the mass of solvent in kg.
2. One mole of water has a mass of 18 g.

Question 23: The molarity of a H2SO4 solution of density 1.84 g cm-3 containing 98% by mass of H2SO4 is
(a) 18 M
(b) 8.14 M
(c) 18.4 M
(d) 4.18 M
Show Answer & Explanation

Answer: (c) 18.4 M

Explanation:
1. 1 L of solution has mass 1840 g, containing 1840 × 0.98 = 1803.2 g H2SO4.
2. Moles = 1803.2 / 98 = 18.4, so molarity = 18.4 M.

Teacher's Note:
1. M = 10 x d x (mass %) / molar mass.
2. Here 10 × 1.84 × 98 / 98 = 18.4.

Question 24: The molarity of a solution containing 2.8% (mass/volume) KOH (K = 39) is
(a) 0.1 M
(b) 0.5 M
(c) 1 M
(d) 0.2 M
Show Answer & Explanation

Answer: (b) 0.5 M

Explanation:
1. 2.8% w/v means 2.8 g in 100 mL, so 28 g in 1 L.
2. Moles = 28 / 56 = 0.5, so molarity = 0.5 M.

Teacher's Note:
1. Scale w/v to one litre, then divide by the molar mass.
2. Molar mass of KOH is 56 g.

Question 25: Arrange in decreasing order the mass of pure NaOH in (i) 50 g of 40% (w/w) NaOH; (ii) 50 mL of 50% (w/v) NaOH; (iii) 50 g of 15 M NaOH solution (d = 1 g mL-1)
(a) i, ii, iii
(b) iii, ii, i
(c) ii, iii, i
(d) iii = ii = i
Show Answer & Explanation

Answer: (b) iii, ii, i

Explanation:
1. (i) 0.40 × 50 = 20 g.
2. (ii) 50% w/v means 50 g per 100 mL, so 25 g in 50 mL.
3. (iii) 50 g at d = 1 is 50 mL = 0.05 L; moles = 15 × 0.05 = 0.75 mol = 30 g.
4. Order: (iii) 30 g > (ii) 25 g > (i) 20 g.

Teacher's Note:
1. Convert every concentration to mass of solute.
2. w/v is grams per 100 mL of solution.

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