JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 18

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: If 1.5 mol of oxygen combine with Al to form \( Al_2O_3 \), the weight of Al used is (Al = 27)
(a) 27 g
(b) 40.5 g
(c) 54 g
(d) 81 g
Show Answer & Explanation

Answer: (c) 54 g

Explanation:
1. \( \ce{4Al + 3O2 -> 2Al2O3} \). Al = 1.5 x 4/3 = 2 mol.
2. Mass = 2 x 27 = 54 g.

Teacher's Note:
1. Balance the equation: 4 mol of Al react with 3 mol of \( \ce{O2} \).

Question 2: The moles of potassium chlorate that need to be heated to produce 11.2 L of oxygen at NTP is
(a) 1/2 mol
(b) 1/3 mol
(c) 1/4 mol
(d) 2/3 mol
Show Answer & Explanation

Answer: (b) 1/3 mol

Explanation:
1. \( \ce{O2} \) = 11.2/22.4 = 0.5 mol.
2. \( \ce{2KClO3 -> 2KCl + 3O2} \): \( \ce{KClO3} \) = 0.5 x 2/3 = 1/3 mol.

Teacher's Note:
1. Two moles of \( \ce{KClO3} \) give three moles of oxygen.

Question 3: In Mond's process, \( \ce{Ni + 4CO -> Ni(CO)4} \). If the CO is obtained by mixing 6 g of carbon with 44 g of \( \ce{CO2} \) (\( \ce{C + CO2 -> 2CO} \)), the mass of Ni consumed is (Ni = 58.7)
(a) 14.675 g
(b) 29 g
(c) 58 g
(d) 28 g
Show Answer & Explanation

Answer: (a) 14.675 g

Explanation:
1. C = 6/12 = 0.5 mol; \( \ce{CO2} \) = 1 mol. Carbon is limiting, so CO = 2 x 0.5 = 1 mol.
2. Ni = 1/4 mol = 0.25 mol.
3. Mass = 0.25 x 58.7 = 14.675 g.

Teacher's Note:
1. Find the limiting reagent in the first step. Four moles of CO are needed for each mole of Ni.

Question 4: For the reaction \( \ce{2P + Q -> R} \), 8 mol of P and 5 mol 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. P/2 = 4 and Q/1 = 5, so P is limiting.
2. R = 4 mol.

Teacher's Note:
1. Divide each amount by its coefficient. The smaller number is the limiting reagent.

Question 5: How many moles of \( \ce{Zn(FeS2)} \) can be made from 2 mol of zinc, 3 mol of iron and 5 mol of sulphur?
(a) 2 mol
(b) 3 mol
(c) 4 mol
(d) 5 mol
Show Answer & Explanation

Answer: (a) 2 mol

Explanation:
1. Each formula unit needs 1 Zn, 1 Fe and 2 S.
2. Zn = 2, Fe = 3, S/2 = 2.5. The smallest is 2, so zinc is limiting.
3. Product = 2 mol.

Teacher's Note:
1. Divide each element by its number in the formula. The smallest value limits the product.

Question 6: 0.5 mol of \( \ce{H2SO4} \) is mixed with 0.2 mol of \( \ce{Ca(OH)2} \). The maximum number of moles of \( \ce{CaSO4} \) formed is
(a) 0.2
(b) 0.5
(c) 0.4
(d) 1.5
Show Answer & Explanation

Answer: (a) 0.2

Explanation:
1. \( \ce{H2SO4 + Ca(OH)2 -> CaSO4 + 2H2O} \) is 1 : 1.
2. \( \ce{Ca(OH)2} \) is limiting, so \( \ce{CaSO4} \) = 0.2 mol.

Teacher's Note:
1. Calcium limits the amount of calcium sulphate.

Question 7: The mass of 70% \( \ce{H2SO4} \) required for neutralisation of 1 mol of NaOH is
(a) 49 g
(b) 98 g
(c) 70 g
(d) 34.3 g
Show Answer & Explanation

Answer: (c) 70 g

Explanation:
1. \( \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} \): \( \ce{H2SO4} \) = 0.5 mol = 49 g (pure).
2. 70% acid: mass of solution = 49/0.70 = 70 g.

Teacher's Note:
1. Find the pure mass first, then divide by the percentage purity.

Question 8: 358 g of \( \ce{TiCl4} \) is reacted with 96 g of Mg (\( \ce{TiCl4 + 2Mg -> Ti + 2MgCl2} \)). The percentage yield of Ti if 32 g of Ti is actually obtained is (Ti = 48, Mg = 24, Cl = 35.5)
(a) 35.38%
(b) 66.6%
(c) 100%
(d) 60%
Show Answer & Explanation

Answer: (a) 35.38%

Explanation:
1. \( \ce{TiCl4} \) = 358/190 = 1.884 mol. Mg = 4 mol (needs only 3.77 mol), so \( \ce{TiCl4} \) is limiting.
2. Theoretical Ti = 1.884 mol x 48 = 90.4 g.
3. Yield = 32/90.4 x 100 = 35.4%.

Teacher's Note:
1. Percentage yield = actual / theoretical x 100, with theoretical based on the limiting reagent.

Question 9: 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 respectively (I = 127, Cl = 35.5)
(a) 0.1 mol, 0.1 mol
(b) 0.1 mol, 0.2 mol
(c) 0.5 mol, 0.5 mol
(d) 0.2 mol, 0.2 mol
Show Answer & Explanation

Answer: (a) 0.1 mol, 0.1 mol

Explanation:
1. I atoms = 0.2 mol. Cl atoms = 14.2/35.5 = 0.4 mol.
2. Let ICl = x, ICl3 = y: x + y = 0.2 and x + 3y = 0.4.
3. y = 0.1 and x = 0.1.

Teacher's Note:
1. Write one atom balance for each element and solve both equations.

Question 10: 0.05 mol of \( \ce{LiAlH4} \) in ether is placed in a flask containing 74 g (1 mol) of t-butyl alcohol. The product \( \ce{LiAlHC12H27O3} \) weighs 12.7 g. If lithium atoms are conserved, the percentage yield is (Li = 7, Al = 27, H = 1, C = 12, O = 16)
(a) 25%
(b) 75%
(c) 100%
(d) 15%
Show Answer & Explanation

Answer: (c) 100%

Explanation:
1. Molar mass of the product = 254 g/mol.
2. Theoretical = 0.05 x 254 = 12.7 g.
3. Yield = 12.7/12.7 x 100 = 100%.

Teacher's Note:
1. Conserve lithium: moles of product equal moles of \( \ce{LiAlH4} \).

Question 11: NX is produced by: M + \( X_2 \) -> \( MX_2 \); 3\( MX_2 \) + \( X_2 \) -> \( M_3X_8 \); \( M_3X_8 \) + \( N_2CO_3 \) -> NX + \( CO_2 \) + \( M_3O_4 \). The amount of M (metal) consumed to produce 206 g of NX is (M = 56, N = 23, X = 80)
(a) 42 g
(b) 56 g
(c) 14/3 g
(d) 7/4 g
Show Answer & Explanation

Answer: (a) 42 g

Explanation:
1. NX = 103 g/mol, so 206 g = 2 mol.
2. Balanced last step: \( M_3X_8 + 4N_2CO_3 -> 8NX + 4CO_2 + M_3O_4 \). 8 mol NX comes from 3 mol M.
3. M = 2 x 3/8 = 0.75 mol = 42 g.

Teacher's Note:
1. Atom conservation: three metal atoms end up in eight NX units.

Question 12: A mixture of \( \ce{CaCl2} \) and NaCl weighing 4.44 g was treated to precipitate all the Ca as \( \ce{CaCO3} \), which on heating gave 1.12 g of CaO. Which statements are correct? (1) The mixture contains 50% NaCl (2) The mixture contains 60% \( \ce{CaCl2} \) (3) Mass of \( \ce{CaCl2} \) is 2.22 g (4) Mass of \( \ce{CaCl2} \) is 1.11 g
(a) 1 and 3 only
(b) 2 and 4 only
(c) 1 and 4 only
(d) 2 and 3 only
Show Answer & Explanation

Answer: (a) 1 and 3 only

Explanation:
1. CaO = 1.12/56 = 0.02 mol = \( \ce{CaCl2} \) 0.02 mol = 2.22 g.
2. NaCl = 4.44 - 2.22 = 2.22 g, which is 50%.
3. So statements 1 and 3 are correct.

Teacher's Note:
1. Calcium is conserved from \( \ce{CaCl2} \) to CaO.

Question 13: In the reaction \( X^- \) + \( XO_3^- \) + \( H^+ \) -> \( X_2 \) + \( H_2O \), the molar ratio in which \( X^- \) and \( XO_3^- \) react is
(a) 1 : 5
(b) 5 : 1
(c) 2 : 3
(d) 3 : 2
Show Answer & Explanation

Answer: (b) 5 : 1

Explanation:
1. X- loses 1 e (-1 to 0). \( XO_3^- \) gains 5 e (+5 to 0).
2. So 5 X- react with 1 \( XO_3^- \): \( \ce{5X- + XO3- + 6H+ -> 3X2 + 3H2O} \).

Teacher's Note:
1. The ratio of the species is the inverse of the ratio of electrons exchanged per species.

Question 14: Equal moles of \( \ce{H2O} \) and NaCl are present in a solution. The molality of the NaCl solution is
(a) 0.55
(b) 55.5
(c) 1.00
(d) 0.18
Show Answer & Explanation

Answer: (b) 55.5

Explanation:
1. 1 mol of water = 18 g = 0.018 kg.
2. Molality = 1/0.018 = 55.5 mol/kg.

Teacher's Note:
1. Molality = moles of solute per kg of solvent.

Question 15: The decreasing order of mass of pure NaOH in each of the aqueous solutions (I) 50 g of 40% (w/w) NaOH, (II) 50 mL of 50% (w/v) NaOH, (III) 50 mL of 15 M NaOH (density 1 g/mL) is
(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) 40% of 50 g = 20 g.
2. (II) 50% w/v means 25 g in 50 mL = 25 g.
3. (III) 15 M x 0.050 L = 0.75 mol = 30 g.
4. Order: III > II > I.

Teacher's Note:
1. Convert each to grams of NaOH.

Question 16: The mole fraction of A in water is 0.2. The molality of A in water is
(a) 13.9
(b) 15.5
(c) 14.5
(d) 16.8
Show Answer & Explanation

Answer: (a) 13.9

Explanation:
1. Take 1 mol of solution: A = 0.2 mol; water = 0.8 mol = 14.4 g.
2. Molality = 0.2/0.0144 = 13.9 mol/kg.

Teacher's Note:
1. Convert the moles of solvent to kg.

Question 17: The molarity of a 2.8% (mass/volume) solution of KOH is (K = 39)
(a) 0.1 M
(b) 0.5 M
(c) 0.2 M
(d) 1 M
Show Answer & Explanation

Answer: (b) 0.5 M

Explanation:
1. 2.8% (w/v) = 2.8 g per 100 mL = 28 g/L.
2. Moles = 28/56 = 0.5 mol/L.

Teacher's Note:
1. % (w/v) x 10 gives g/L.

Question 18: A solution of \( \ce{FeCl3} \) is M/30 in concentration. Its molarity for the \( Cl^- \) ion is
(a) M/90
(b) M/30
(c) M/10
(d) M/5
Show Answer & Explanation

Answer: (c) M/10

Explanation:
1. Each \( \ce{FeCl3} \) gives 3 Cl-.
2. Concentration of Cl- = 3 x M/30 = M/10.

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

Question 19: If 500 mL of 1 M glucose solution is mixed with 500 mL of 1 M glucose solution, the final molarity is
(a) 1 M
(b) 0.5 M
(c) 2 M
(d) 1.5 M
Show Answer & Explanation

Answer: (a) 1 M

Explanation:
1. Total moles = 1 mol in 1 L.
2. Molarity = 1 M. Mixing equal concentrations does not change the concentration.

Teacher's Note:
1. Add moles and add volumes.

Question 20: The volume of water to be added to a mixture of 250 mL of 0.6 M HCl and 750 mL of 0.2 M HCl to obtain a 0.25 M solution is
(a) 750 mL
(b) 100 mL
(c) 200 mL
(d) 300 mL
Show Answer & Explanation

Answer: (c) 200 mL

Explanation:
1. Moles = 0.25 x 0.6 + 0.75 x 0.2 = 0.15 + 0.15 = 0.30 mol.
2. Final volume for 0.25 M = 0.30/0.25 = 1.2 L = 1200 mL.
3. Water = 1200 - 1000 = 200 mL.

Teacher's Note:
1. Moles stay constant on dilution.

Question 21: The volume of a 0.8 M solution that contains 100 millimoles of the solute is
(a) 100 mL
(b) 125 mL
(c) 500 mL
(d) 62.5 mL
Show Answer & Explanation

Answer: (b) 125 mL

Explanation:
1. Volume = moles / molarity = 0.100/0.8 = 0.125 L = 125 mL.

Teacher's Note:
1. Volume = moles / molarity.

Question 22: 2 M of 100 mL \( Na_2SO_4 \) is mixed with 3 M of 100 mL NaCl and 1 M of 200 mL \( CaCl_2 \) solution. The ratio of the concentration of cations to anions is
(a) 1/2
(b) 2
(c) 1.5
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Cations: Na+ = 0.4 + 0.3 = 0.7 mol; Ca2+ = 0.2 mol; total = 0.9 mol.
2. Anions: sulphate = 0.2 mol; Cl- = 0.3 + 0.4 = 0.7 mol; total = 0.9 mol.
3. Ratio = 1.

Teacher's Note:
1. Count the moles of each ion. The solution is electrically neutral in terms of equivalents.

Question 23: Assertion (A): The weight percentage of a compound A in a solution is (mass of A / total mass of solution) x 100.
Reason (R): The mole fraction of a component A is (number of moles of A) / (total number of moles of all components).
(a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.
(b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) Assertion is false but the Reason is true.
Show Answer & Explanation

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. Both formulas are correct.
2. The mole fraction definition does not explain the weight percentage definition.

Teacher's Note:
1. Different concentration terms are independent definitions.

Question 24: Assertion (A): The ratio of the mass of 100 billion atoms of magnesium to the mass of 100 billion atoms of lead is 24/207.
Reason (R): Atomic weights are relative masses.
(a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.
(b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) Assertion is false but the Reason is true.
Show Answer & Explanation

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Equal numbers of atoms have masses in the ratio of their atomic masses, 24 : 207.
2. This works because atomic weights are relative masses.

Teacher's Note:
1. For equal numbers of atoms, the mass ratio equals the atomic mass ratio.

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Some Basic Concepts Of Chemistry Objective Questions & Solutions for JEE Chemistry

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