JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 17

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

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Multiple Choice Questions

Question 1: The mass of iron converted into its oxide by 18 g of steam in the reaction \( \ce{Fe + H2O -> Fe3O4 + H2} \) is (Fe = 56)
(a) 42 g
(b) 56 g
(c) 21 g
(d) 84 g
Show Answer & Explanation

Answer: (a) 42 g

Explanation:
1. Balanced: \( \ce{3Fe + 4H2O -> Fe3O4 + 4H2} \).
2. Steam = 18/18 = 1 mol, so Fe = 3/4 mol.
3. Mass = 0.75 x 56 = 42 g.

Teacher's Note:
1. Balance the equation first, then use the 3 : 4 mole ratio.

Question 2: A sample of \( \ce{KClO3} \) on decomposition gives 448 mL of oxygen at NTP. The weight of oxygen, the weight of \( \ce{KClO3} \) originally taken and the weight of KCl formed are respectively (K = 39, Cl = 35.5, O = 16)
(a) 0.64 g, 1.63 g and 0.99 g
(b) 0.64 g, 0.99 g and 1.63 g
(c) 0.32 g, 1.63 g and 0.99 g
(d) 0.64 g, 3.27 g and 1.99 g
Show Answer & Explanation

Answer: (a) 0.64 g, 1.63 g and 0.99 g

Explanation:
1. \( \ce{2KClO3 -> 2KCl + 3O2} \). \( \ce{O2} \) = 0.448/22.4 = 0.02 mol = 0.64 g.
2. \( \ce{KClO3} \) = 0.02 x 2/3 = 0.01333 mol = 1.63 g.
3. KCl = 0.01333 mol x 74.5 = 0.99 g.

Teacher's Note:
1. Work from moles of oxygen using the 2 : 3 ratio. Check: 1.63 - 0.64 = 0.99 g.

Question 3: 50 g of \( \ce{CaCO3} \) reacts with 73.5 g of \( \ce{H3PO4} \): \( \ce{3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3H2O + 3CO2} \). The moles of \( \ce{Ca3(PO4)2} \) formed and of reagent left unreacted are respectively
(a) 0.25 mol and 0.5 mol
(b) 0.375 mol and 0.0 mol
(c) 0.167 mol and 0.167 mol
(d) 0.167 mol and 0.417 mol
Show Answer & Explanation

Answer: (d) 0.167 mol and 0.417 mol

Explanation:
1. \( \ce{CaCO3} \) = 0.5 mol; \( \ce{H3PO4} \) = 73.5/98 = 0.75 mol.
2. 0.5/3 = 0.167 and 0.75/2 = 0.375, so \( \ce{CaCO3} \) is limiting.
3. \( \ce{Ca3(PO4)2} \) = 0.167 mol. \( \ce{H3PO4} \) used = 0.333 mol, left = 0.417 mol.

Teacher's Note:
1. Find the limiting reagent first, then find the leftover of the other reagent.

Question 4: The percentage yield of \( \ce{Br2 + Cl2 -> 2BrCl} \) in \( \ce{CCl4} \) is 80%. From 0.025 mol of \( \ce{Br2} \) and 0.025 mol of \( \ce{Cl2} \), the amount of BrCl formed and the amount of \( \ce{Br2} \) left unchanged are respectively
(a) 0.040 mol and 0.005 mol
(b) 0.050 mol and 0.005 mol
(c) 0.040 mol and 0.020 mol
(d) 0.020 mol and 0.005 mol
Show Answer & Explanation

Answer: (a) 0.040 mol and 0.005 mol

Explanation:
1. Theoretical BrCl = 2 x 0.025 = 0.050 mol.
2. Actual BrCl = 0.80 x 0.050 = 0.040 mol, which needs 0.020 mol of each reactant.
3. \( \ce{Br2} \) left = 0.025 - 0.020 = 0.005 mol.

Teacher's Note:
1. Actual product = yield x theoretical. Reactant consumed corresponds to the actual product.

Question 5: \( \ce{KClO3} \) decomposes in two parallel reactions: (i) \( \ce{2KClO3 -> 2KCl + 3O2} \) and (ii) \( \ce{4KClO3 -> 3KClO4 + KCl} \). If 3 mol of \( \ce{O2} \) and 1 mol of \( \ce{KClO4} \) are produced along with other products, the initial moles of \( \ce{KClO3} \) are
(a) 10/3 mol
(b) 3 mol
(c) 4 mol
(d) 2 mol
Show Answer & Explanation

Answer: (a) 10/3 mol

Explanation:
1. Reaction (i): 3 mol \( \ce{O2} \) needs 2 mol \( \ce{KClO3} \).
2. Reaction (ii): 1 mol \( \ce{KClO4} \) needs 4/3 mol \( \ce{KClO3} \).
3. Total = 2 + 4/3 = 10/3 mol.

Teacher's Note:
1. For parallel reactions, add the reactant consumed in each branch.

Question 6: The mass of KOH needed to prepare 100 mL of 1 M KOH solution is (K = 39)
(a) 56 g
(b) 0.56 g
(c) 5.6 g
(d) 11.2 g
Show Answer & Explanation

Answer: (c) 5.6 g

Explanation:
1. Moles = 1 x 0.100 = 0.1 mol.
2. Mass = 0.1 x 56 = 5.6 g.

Teacher's Note:
1. Mass = molarity x volume in litre x molar mass.

Question 7: The mass of \( \ce{NaNO3} \) to be weighed to make 50 mL of an aqueous solution containing 70 mg of \( Na^+ \) per mL is (Na = 23, N = 14, O = 16)
(a) 3.5 g
(b) 6.5 g
(c) 12.9 g
(d) 25.9 g
Show Answer & Explanation

Answer: (c) 12.9 g

Explanation:
1. Na+ needed = 70 x 50 = 3500 mg = 3.5 g = 0.152 mol.
2. \( \ce{NaNO3} \) = 0.152 mol x 85 = 12.9 g.

Teacher's Note:
1. One \( \ce{NaNO3} \) gives one Na+ ion, so moles of salt equal moles of sodium.

Question 8: A 2 M NaOH solution has density 1 g/mL. The mole fraction of NaOH is
(a) 0.0400
(b) 0.0200
(c) 0.0377
(d) 0.0817
Show Answer & Explanation

Answer: (c) 0.0377

Explanation:
1. NaOH = 2 mol; water = 920/18 = 51.1 mol.
2. Mole fraction = 2/(2 + 51.1) = 0.0377.

Teacher's Note:
1. Mole fraction = moles of solute / total moles.

Question 9: A 5 m NaOH solution has density 1.5 g/mL. The mole fraction of NaOH is
(a) 0.0826
(b) 0.0500
(c) 0.0377
(d) 0.1667
Show Answer & Explanation

Answer: (a) 0.0826

Explanation:
1. Per 1 kg of water: NaOH = 5 mol; water = 1000/18 = 55.6 mol.
2. Mole fraction = 5/(5 + 55.6) = 0.0826.

Teacher's Note:
1. Mole fraction is independent of density.

Question 10: A 5 m NaOH solution has density 1.5 g/mL. The percentage (w/v) of NaOH in it is
(a) 16.7%
(b) 20%
(c) 33.3%
(d) 25%
Show Answer & Explanation

Answer: (d) 25%

Explanation:
1. Per 1 kg of water: NaOH = 200 g in 1200 g of solution (800 mL).
2. % (w/v) = 200/800 x 100 = 25%. (The % w/w is 16.7%.)

Teacher's Note:
1. % (w/v) is grams of solute per 100 mL of solution, so volume from density is needed.

Question 11: 300 g of 30% (w/w) NaOH solution is mixed with 500 g of 40% (w/w) NaOH solution. The mass percentage of the final solution is
(a) 36.25%
(b) 35%
(c) 37.5%
(d) 34.5%
Show Answer & Explanation

Answer: (a) 36.25%

Explanation:
1. NaOH = 0.30 x 300 + 0.40 x 500 = 90 + 200 = 290 g.
2. Total mass = 800 g.
3. Mass percent = 290/800 x 100 = 36.25%.

Teacher's Note:
1. Mass is additive in mixing, so add masses of solute and solution.

Question 12: The concentration of chloride ion in a solution made by mixing 1 mol each of \( BaCl_2 \), NaCl and HCl in 500 mL of water is
(a) 8 M
(b) 4 M
(c) 6 M
(d) 3 M
Show Answer & Explanation

Answer: (a) 8 M

Explanation:
1. Cl- moles = 2 (from \( \ce{BaCl2} \)) + 1 + 1 = 4 mol.
2. Concentration = 4/0.5 = 8 M.

Teacher's Note:
1. Count every chloride ion that each salt releases.

Question 13: The volume of water to be added to 50 mL of \( \ce{HNO3} \) (density 1.5 g/mL, 63.0% by weight) to get a 1 M solution is (assume volumes are additive)
(a) 700 mL
(b) 750 mL
(c) 650 mL
(d) 1000 mL
Show Answer & Explanation

Answer: (a) 700 mL

Explanation:
1. Mass of solution = 75 g; \( \ce{HNO3} \) = 47.25 g = 0.75 mol.
2. For 1 M, final volume = 0.75 L = 750 mL.
3. Water to add = 750 - 50 = 700 mL.

Teacher's Note:
1. Moles stay the same on dilution. Find the final volume and subtract the original.

Question 14: The maximum volume of 3 M KOH solution that can be prepared from 1 L each of 1 M KOH and 6 M KOH solutions by using water is
(a) 7/3 L
(b) 2 L
(c) 3 L
(d) 1 L
Show Answer & Explanation

Answer: (a) 7/3 L

Explanation:
1. Total KOH available = 1 x 1 + 1 x 6 = 7 mol.
2. Volume of 3 M solution = 7/3 L = 2.33 L (about 0.33 L of water is needed in addition to the 2 L of solutions).

Teacher's Note:
1. Use all of both solutions to get the largest amount of solute, then dilute to 3 M.

Question 15: Which of the following contains the greatest number of atoms?
(a) 1.0 g of butane, \( \ce{C4H10} \)
(b) 1.0 g of nitrogen, \( \ce{N2} \)
(c) 1.0 g of silver, Ag
(d) 1.0 g of water, \( \ce{H2O} \)
Show Answer & Explanation

Answer: (a) 1.0 g of butane, \( \ce{C4H10} \)

Explanation:
1. Moles of atoms: butane = 14/58 = 0.241; \( \ce{N2} \) = 2/28 = 0.071; Ag = 1/108 = 0.0093; water = 3/18 = 0.167.
2. Butane has the most atoms.

Teacher's Note:
1. Multiply moles of the substance by atoms per molecule, then compare.

Question 16: Four 1-litre flasks are separately filled with the gases \( \ce{H2} \), He, \( \ce{O2} \) and \( \ce{O3} \) at the same temperature and pressure. The ratio of the total number of atoms of these gases present in the different flasks is
(a) 1 : 1 : 1 : 1
(b) 1 : 2 : 2 : 3
(c) 2 : 1 : 2 : 3
(d) 3 : 2 : 2 : 1
Show Answer & Explanation

Answer: (c) 2 : 1 : 2 : 3

Explanation:
1. By Avogadro's law, the flasks hold equal numbers of molecules.
2. Atoms per molecule: \( \ce{H2} \) = 2, He = 1, \( \ce{O2} \) = 2, \( \ce{O3} \) = 3.
3. Ratio of atoms = 2 : 1 : 2 : 3.

Teacher's Note:
1. Same volume, T and P means the same number of molecules. Then multiply by atomicity.

Question 17: 3 g of a hydrocarbon on combustion in excess oxygen produces 8.8 g of \( \ce{CO2} \) and 5.4 g of \( \ce{H2O} \). The data illustrate the law of
(a) conservation of mass
(b) multiple proportions
(c) constant proportions
(d) none of these
Show Answer & Explanation

Answer: (a) conservation of mass

Explanation:
1. C in 8.8 g \( \ce{CO2} \) = 2.4 g. H in 5.4 g \( \ce{H2O} \) = 0.6 g.
2. 2.4 + 0.6 = 3.0 g, equal to the mass of hydrocarbon burnt.
3. The mass of the elements in the products equals the mass of the reactant: conservation of mass.

Teacher's Note:
1. Account for the elements in the products. If the masses add up to the reactant, mass is conserved.

Question 18: Boron has two stable isotopes, \( ^{10}B \) (relative abundance 19%) and \( ^{11}B \) (81%). The atomic mass that should appear for boron in the periodic table is
(a) 10.8
(b) 10.2
(c) 11.2
(d) 10.6
Show Answer & Explanation

Answer: (a) 10.8

Explanation:
1. Average = (10 x 19 + 11 x 81)/100 = (190 + 891)/100 = 10.81.

Teacher's Note:
1. Average atomic mass is the abundance-weighted mean.

Question 19: The empirical formula of a compound of molecular mass 120 is \( \ce{CH2O} \). The molecular formula of the compound is
(a) \( \ce{C2H4O2} \)
(b) \( \ce{C4H8O4} \)
(c) \( \ce{C3H6O3} \)
(d) all of these
Show Answer & Explanation

Answer: (b) \( \ce{C4H8O4} \)

Explanation:
1. Empirical formula mass = 30.
2. n = 120/30 = 4, so molecular formula = \( \ce{C4H8O4} \).

Teacher's Note:
1. Molecular formula = (empirical formula) x n, where n = molar mass / empirical mass.

Question 20: A compound contains 20% Ca and 80% Br by weight and has molecular weight 200. Its molecular formula is (Ca = 40, Br = 80)
(a) \( Ca_{1/2}Br \)
(b) \( CaBr_2 \)
(c) CaBr
(d) \( Ca_2Br \)
Show Answer & Explanation

Answer: (b) \( CaBr_2 \)

Explanation:
1. Moles: Ca = 20/40 = 0.5; Br = 80/80 = 1. Ratio 1 : 2 so empirical formula \( \ce{CaBr2} \), mass 200.
2. Molar mass = 200, so n = 1 and the formula is \( \ce{CaBr2} \).

Teacher's Note:
1. Find the empirical formula and compare its mass with the molar mass.

Question 21: A compound possesses 8% sulphur by mass. The least molecular mass is (S = 32)
(a) 200
(b) 400
(c) 155
(d) 355
Show Answer & Explanation

Answer: (b) 400

Explanation:
1. Minimum molar mass has one S atom: 32/0.08 = 400.

Teacher's Note:
1. If the compound has at least one atom of the element, the least molar mass = atomic mass / mass fraction.

Question 22: Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is 69.98%. Its molar mass is (C = 12)
(a) 176.5
(b) 252.2
(c) 287.6
(d) 360.1
Show Answer & Explanation

Answer: (d) 360.1

Explanation:
1. Mass of C per mole = 21 x 12 = 252 g.
2. Molar mass = 252/0.6998 = 360.1 g/mol.

Teacher's Note:
1. Mass of the element per mole divided by its mass fraction gives the molar mass.

Question 23: 12 g of an alkaline earth metal gives 14.8 g of its nitride. The atomic weight of the metal is (N = 14)
(a) 12
(b) 20
(c) 40
(d) 14.8
Show Answer & Explanation

Answer: (c) 40

Explanation:
1. Nitride is \( M_3N_2 \). Mass of N = 14.8 - 12 = 2.8 g = 0.2 mol.
2. Metal = 3/2 x 0.2 = 0.3 mol.
3. Atomic weight = 12/0.3 = 40.

Teacher's Note:
1. Use the formula \( M_3N_2 \): three metal atoms for every two N atoms.

Question 24: For the reaction \( \ce{2P + Q -> R} \), 8 mol of P and excess of Q will produce
(a) 8 mol of R
(b) 5 mol of R
(c) 4 mol of R
(d) 13 mol of R
Show Answer & Explanation

Answer: (c) 4 mol of R

Explanation:
1. Two moles of P give one mole of R, so R = 8/2 = 4 mol.

Teacher's Note:
1. The coefficient of P is 2, so halve the moles.

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

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