JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 33

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

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Multiple Choice Questions

Question 1: X and Y form X2Y3 and X3Y4. If 0.20 mol of X2Y3 weighs 32.0 g and 0.4 mol of X3Y4 weighs 92.8 g, the atomic masses of X and Y are respectively
(a) 16.0 and 56.0
(b) 8.0 and 28.0
(c) 28.0 and 8.0
(d) 56.0 and 16.0
Show Answer & Explanation

Answer: (d) 56.0 and 16.0

Explanation:
1. Molar masses: X2Y3 = 160 g mol-1; X3Y4 = 232 g mol-1.
2. 2x + 3y = 160 and 3x + 4y = 232. Solving gives x = 56 and y = 16.

Teacher's Note:
1. Two compounds give two simultaneous equations.
2. These are the atomic masses of Fe and O.

Question 2: In the reaction 4A + 2B + 3C → A4B2C3, the number of moles of product formed from 2 mol A, 1.2 mol B and 1.44 mol C is
(a) 0.5
(b) 0.6
(c) 4.64
(d) 0.48
Show Answer & Explanation

Answer: (d) 0.48

Explanation:
1. Possible product moles: A = 2/4 = 0.5; B = 1.2/2 = 0.6; C = 1.44/3 = 0.48.
2. C gives the smallest value, so it is limiting and 0.48 mol forms.

Teacher's Note:
1. Divide each moles by its coefficient.
2. The smallest quotient identifies the limiting reagent.

Question 3: Which of the following equations is balanced?
(a) 5BiO3- + 22H+ + Mn2+ → 5Bi3+ + 7H2O + MnO4-
(b) 5BiO3- + 14H+ + 2Mn2+ → 5Bi3+ + 7H2O + 2MnO4-
(c) 2BiO3- + 4H+ + Mn2+ → 2Bi3+ + 2H2O + MnO4-
(d) 6BiO3- + 12H+ + 3Mn2+ → 6Bi3+ + 6H2O + 3MnO4-
Show Answer & Explanation

Answer: (b) 5BiO3- + 14H+ + 2Mn2+ → 5Bi3+ + 7H2O + 2MnO4-

Explanation:
1. Electron balance: Bi goes from +5 to +3 (2 e- each) and Mn from +2 to +7 (5 e- each), so 5 Bi to 2 Mn.
2. Option (b): charge on the left = -5 + 14 + 4 = +13; on the right = +15 - 2 = +13. O: 15 on the left; 7 + 8 = 15 on the right. H: 14 on both sides.
3. The other options fail the charge, electron or atom balance.

Teacher's Note:
1. Check electrons, charge and each atom.
2. Balance electrons first, then H+ and water.

Question 4: When iodine disproportionates to iodide and iodate in alkaline medium, the ratio of iodate to iodide ions formed is
(a) 5 : 1
(b) 1 : 5
(c) 1 : 3
(d) 3 : 1
Show Answer & Explanation

Answer: (b) 1 : 5

Explanation:
1. 3I2 + 6OH- → 5I- + IO3- + 3H2O.
2. Iodate : iodide = 1 : 5.

Teacher's Note:
1. In disproportionation the electrons gained equal the electrons lost.
2. One I goes 0 to +5 (5 e-) and five I go 0 to -1.

Question 5: The temperature at which the molarity of pure water is equal to its molality is
(a) 273 K
(b) 298 K
(c) 277 K
(d) none of these
Show Answer & Explanation

Answer: (c) 277 K

Explanation:
1. Molarity = 1000 d / 18 and molality = 1000 / 18 for pure water.
2. They are equal only when d = 1 g mL-1, which is at 4 °C (about 277 K).

Teacher's Note:
1. Water has maximum density at 4 °C.
2. Molarity depends on volume, which depends on temperature.

Question 6: 5.85 g of NaCl is dissolved in 1 L of pure water. The number of ions in 1 mL of this solution is (take volume as 1 L)
(a) 6.02 × 1019
(b) 1.2 × 1020
(c) 6.02 × 1020
(d) 1.2 × 1022
Show Answer & Explanation

Answer: (b) 1.2 × 1020

Explanation:
1. Concentration = 5.85/58.5 = 0.1 M, so 1 mL contains 10-4 mol NaCl.
2. Ions = 2 × 10-4 × 6.02 × 1023 = 1.2 × 1020.

Teacher's Note:
1. Each NaCl gives two ions.
2. 1 mL is one thousandth of a litre.

Question 7: The correct relation between molality (m), molarity (M), density of solution (d, g mL-1) and molar mass of solute (M_2) is
(a) m = M / (d + M M_2) × 1000
(b) m = M / (1000d - M M_2) × 1000
(c) m = (d + M M_2) / M × 1000
(d) m = (1000d - M M_2) / M × 1000
Show Answer & Explanation

Answer: (b) m = M / (1000d - M M_2) × 1000

Explanation:
1. In 1 L of solution: moles of solute = M; mass of solution = 1000d; mass of solute = M M_2.
2. Mass of solvent = (1000d - M M_2) g, so m = M / (1000d - M M_2) × 1000.

Teacher's Note:
1. Derive by taking 1 L of solution.
2. Molality uses solvent mass in kg.

Question 8: The volume of O2 needed at STP for the complete combustion of 1 kg of carbon (C + O2 → CO2) is
(a) 1866.67 L
(b) 933.33 L
(c) 4666.67 L
(d) 3733.33 L
Show Answer & Explanation

Answer: (a) 1866.67 L

Explanation:
1. Moles of C = 1000/12 = 83.33 mol.
2. O2 needed = 83.33 mol, so volume = 83.33 × 22.4 = 1866.67 L.

Teacher's Note:
1. The ratio C : O2 is 1 : 1.
2. Convert grams to moles first.

Question 9: In 6Li + N2 → 2Li3N, the mass of Li3N that can be prepared from 3.5 g of Li and 8.4 g of N2 is (Li = 7, N = 14)
(a) 21.00 g
(b) 2.91 g
(c) 5.83 g
(d) 10.50 g
Show Answer & Explanation

Answer: (c) 5.83 g

Explanation:
1. Li = 3.5/7 = 0.5 mol; N2 = 8.4/28 = 0.3 mol.
2. Li needs 0.5/6 = 0.0833 of the reaction; N2 allows 0.3, so Li is limiting.
3. Li3N = 0.5 × 2/6 = 0.1667 mol = 0.1667 × 35 = 5.83 g.

Teacher's Note:
1. Compare moles / coefficient for each reactant.
2. The limiting reagent decides the yield.

Question 10: A vessel contains 0.15 mol KO2 and 0.10 mol H2O: 2KO2 + 2H2O → 2KOH + H2O2 + O2. The limiting reactant and the moles of O2 produced are
(a) H2O is limiting, 0.05 mol O2
(b) KO2 is limiting, 0.05 mol O2
(c) KO2 is limiting, 0.075 mol O2
(d) H2O is limiting, 0.075 mol O2
Show Answer & Explanation

Answer: (a) H2O is limiting, 0.05 mol O2

Explanation:
1. The ratio KO2 : H2O is 1 : 1, so the smaller amount (0.10 mol H2O) limits the reaction.
2. 2 mol H2O give 1 mol O2, so 0.10 mol gives 0.05 mol O2.

Teacher's Note:
1. Both reactants have coefficient 2.
2. Use the limiting reagent for the product.

Question 11: Equal masses of mercury and iodine react completely to give a mixture of mercurous iodide (Hg2I2) and mercuric iodide (HgI2), leaving no reactant. The ratio of the masses of Hg2I2 to HgI2 is (Hg = 200.6, I = 126.9)
(a) 1 : 0.523
(b) 1 : 0.653
(c) 0.523 : 1
(d) 0.732 : 1
Show Answer & Explanation

Answer: (c) 0.523 : 1

Explanation:
1. Let a mol Hg2I2 and b mol HgI2 form. Hg = 2a + b and I2 = a + b moles.
2. Equal masses: (2a + b) × 200.6 = (a + b) × 253.8, which gives a = 0.361 b.
3. Mass ratio = 0.361 × 654.0 / 454.4 = 0.52, that is 0.523 : 1 to the accuracy of the atomic masses used.

Teacher's Note:
1. Use equal masses to relate the two unknowns.
2. Compare the answer with the options to within rounding.

Question 12: 2.7 g of aluminium is heated with 75.0 mL of H2SO4 (specific gravity 1.2, 25% by mass). After the metal dissolves, the solution is diluted to 400 mL. The molarity of the free H2SO4 is
(a) 0.560 M
(b) 0.198 M
(c) 1.056 M
(d) 0.312 M
Show Answer & Explanation

Answer: (b) 0.198 M

Explanation:
1. H2SO4 initially = 75 × 1.2 × 0.25 / 98 = 0.2296 mol; Al = 0.1 mol.
2. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 uses 0.15 mol H2SO4.
3. Free H2SO4 = 0.0796 mol in 0.4 L = 0.199 M.

Teacher's Note:
1. Subtract the acid consumed from the acid taken.
2. Divide by the final volume.

Question 13: The number of Al3+ ions in 100 mL of 0.15 M Al2(SO4)3 solution (density 1.5 g mL-1) is
(a) 1.8 × 1025
(b) 6 × 1022
(c) 1.8 × 1023
(d) 1.8 × 1022
Show Answer & Explanation

Answer: (d) 1.8 × 1022

Explanation:
1. Moles of Al2(SO4)3 = 0.15 × 0.1 = 0.015 mol, containing 0.03 mol Al3+.
2. Ions = 0.03 × 6.02 × 1023 = 1.8 × 1022.

Teacher's Note:
1. Density is not needed for this question.
2. Count two Al3+ per formula unit.

Question 14: Calcium orthophosphate is reduced by magnesium: Ca3(PO4)2 + 8Mg → Ca3P2 + 8MgO. The Ca3P2 is hydrolysed to PH3, burnt to P2O5, and MgO + P2O5 → Mg(PO3)2. The mass of Mg(PO3)2 obtained from 19.2 g of Mg is (Mg = 24, P = 31)
(a) 50.4 g
(b) 32 g
(c) 145.8 g
(d) 18.2 g
Show Answer & Explanation

Answer: (d) 18.2 g

Explanation:
1. Mg = 19.2/24 = 0.8 mol gives 0.1 mol Ca3P2, hence 0.2 mol PH3 and 0.1 mol P2O5; MgO = 0.8 mol.
2. MgO and P2O5 combine 1 : 1, so P2O5 (0.1 mol) is limiting and 0.1 mol Mg(PO3)2 forms.
3. Molar mass of Mg(PO3)2 = 24 + 62 + 96 = 182 g, so mass = 18.2 g.

Teacher's Note:
1. Follow phosphorus atoms through the sequence.
2. Check which of MgO and P2O5 is limiting in the last step.

Question 15: A 10.0 g mixture of CaCl2 and NaCl is treated with Na2CO3 solution. The CaCO3 is heated to CaO, with final mass 1.62 g. The mass percentage of CaCl2 in the mixture is
(a) 16.1%
(b) 11.0%
(c) 32.1%
(d) 21.8%
Show Answer & Explanation

Answer: (c) 32.1%

Explanation:
1. CaO = 1.62/56 = 0.0289 mol = CaCl2.
2. Mass of CaCl2 = 0.0289 × 111 = 3.21 g, which is 32.1% of 10 g.

Teacher's Note:
1. Ca atoms are conserved.
2. Percentage = mass / 10 g × 100.

Question 16: The minimum mass of Ag2CO3 (Ag = 108) required to give enough oxygen for the complete combustion of C2H2 that produces 1.12 L of CO2 at STP is (Ag2CO3 → 2Ag + CO2 + 1/2 O2; C2H2 + 5/2 O2 → 2CO2 + H2O)
(a) 17.25 g
(b) 34.5 g
(c) 138 g
(d) 69 g
Show Answer & Explanation

Answer: (b) 34.5 g

Explanation:
1. CO2 = 1.12/22.4 = 0.05 mol, so C2H2 = 0.025 mol.
2. O2 needed = 0.025 × 5/2 = 0.0625 mol.
3. Each mole of Ag2CO3 gives 1/2 mol O2, so Ag2CO3 = 0.125 mol = 0.125 × 276 = 34.5 g.

Teacher's Note:
1. Work backwards from CO2 to C2H2 to O2 to Ag2CO3.
2. Molar mass of Ag2CO3 is 276 g.

Question 17: 64 g of a mixture of NaCl and KCl was treated with concentrated H2SO4, giving 76 g of mixed sulphates (Na2SO4 and K2SO4). The mass percentage of NaCl in the mixture is about
(a) 57%
(b) 30%
(c) 70%
(d) 43%
Show Answer & Explanation

Answer: (d) 43%

Explanation:
1. Let NaCl = x g. 2NaCl → Na2SO4: 117 g gives 142 g; 2KCl → K2SO4: 149 g gives 174 g.
2. x × 142/117 + (64 - x) × 174/149 = 76 gives x = 27.5 g.
3. Percentage NaCl = 27.5/64 × 100 = 43%.

Teacher's Note:
1. Write one mass balance for the sulphates.
2. Convert the mass of NaCl to a percentage of 64 g.

Question 18: 5C + 2SO2 → CS2 + 4CO. How much CS2 can be produced from 440 kg of waste SO2 with 60 kg of coke if the SO2 conversion is 80%? (C = 12, S = 32)
(a) 38 kg
(b) 95 kg
(c) 19 kg
(d) 76 kg
Show Answer & Explanation

Answer: (d) 76 kg

Explanation:
1. SO2 available for reaction = 0.80 × 440 = 352 kg = 5500 mol; C = 60 000/12 = 5000 mol.
2. C needed for 5500 mol SO2 is 13 750 mol, so carbon is limiting: CS2 = 5000/5 = 1000 mol.
3. Mass = 1000 × 76 g = 76 kg.

Teacher's Note:
1. Compare the moles of the two reactants after applying the conversion.
2. Carbon is limiting here.

Question 19: 358 g of TiCl4 reacts with 96 g of Mg to form Ti (TiCl4 + 2Mg → Ti + 2MgCl2). If 32 g of Ti is actually obtained, the percentage yield is (Ti = 48, Cl = 35.5, Mg = 24)
(a) 60%
(b) 100%
(c) 35.38%
(d) 66.6%
Show Answer & Explanation

Answer: (c) 35.38%

Explanation:
1. TiCl4 = 358/190 = 1.884 mol; Mg = 4 mol, which is more than the 3.77 mol needed, so TiCl4 is limiting.
2. Theoretical Ti = 1.884 × 48 = 90.4 g.
3. Yield = 32/90.4 × 100 = 35.4%.

Teacher's Note:
1. The limiting reagent decides the theoretical yield.
2. Percentage yield = actual / theoretical × 100.

Question 20: In an organic compound of molar mass 108 g mol-1, C, H and N are present in the mass ratio 9 : 1 : 3.5. The molecular formula can be
(a) C4H18N3
(b) C6H8N2
(c) C7H10N
(d) C5H6N3
Show Answer & Explanation

Answer: (b) C6H8N2

Explanation:
1. Moles ratio: C = 9/12 = 0.75, H = 1, N = 3.5/14 = 0.25, so C : H : N = 3 : 4 : 1.
2. Empirical formula C3H4N has mass 54; n = 108/54 = 2.
3. Molecular formula = C6H8N2.

Teacher's Note:
1. Check the mass of each option as well.
2. Divide by the smallest number of moles.

Question 21: Which of the following concentration terms is affected by a change in temperature?
(a) Mole fraction
(b) Weight fraction
(c) Molality
(d) Molarity
Show Answer & Explanation

Answer: (d) Molarity

Explanation:
1. Molarity is moles per unit volume, and volume changes with temperature.
2. Molality, mole fraction and weight fraction depend only on masses or moles.

Teacher's Note:
1. Volume-based terms depend on temperature.
2. Mass-based terms do not.

Question 22: The volume of hydrogen at 273 K and 1 atm consumed in obtaining 21.6 g of boron (atomic mass 10.8) by reducing boron trichloride with hydrogen is
(a) 44.8 L
(b) 67.2 L
(c) 89.6 L
(d) 22.4 L
Show Answer & Explanation

Answer: (b) 67.2 L

Explanation:
1. 2BCl3 + 3H2 → 2B + 6HCl; 21.6 g B = 2 mol.
2. H2 needed = 2 × 3/2 = 3 mol = 3 × 22.4 = 67.2 L.

Teacher's Note:
1. Use the mole ratio B : H2 = 2 : 3.
2. 1 mol of gas at 273 K and 1 atm occupies 22.4 L.

Question 23: 480 mL of a 1.5 M solution of a non-electrolyte is mixed with 520 mL of a 1.2 M solution of the same substance. The molarity of the final mixture is
(a) 1.344 M
(b) 1.50 M
(c) 1.20 M
(d) 2.70 M
Show Answer & Explanation

Answer: (a) 1.344 M

Explanation:
1. Moles = 0.480 × 1.5 + 0.520 × 1.2 = 0.720 + 0.624 = 1.344 mol.
2. Volume = 1.000 L, so molarity = 1.344 M.

Teacher's Note:
1. Add moles and add volumes.
2. Check the answer lies between 1.2 and 1.5.

Question 24: How many moles of magnesium phosphate, Mg3(PO4)2, contain 0.25 mol of oxygen atoms?
(a) 0.02
(b) 2.5 × 10-2
(c) 1.25 × 10-2
(d) 3.125 × 10-2
Show Answer & Explanation

Answer: (d) 3.125 × 10-2

Explanation:
1. One formula unit contains 8 oxygen atoms.
2. Moles = 0.25/8 = 3.125 × 10-2.

Teacher's Note:
1. Divide the moles of atoms by the subscript.
2. Do not forget the 2 outside the bracket multiplies the 4.

Question 25: The density of a 2.05 M solution of acetic acid in water is 1.02 g mL-1. Its molality is
(a) 2.28 mol kg-1
(b) 3.28 mol kg-1
(c) 0.44 mol kg-1
(d) 1.14 mol kg-1
Show Answer & Explanation

Answer: (a) 2.28 mol kg-1

Explanation:
1. 1 L of solution = 1020 g; acetic acid = 2.05 × 60 = 123 g.
2. Water = 1020 - 123 = 897 g.
3. Molality = 2.05/0.897 = 2.28 mol kg-1.

Teacher's Note:
1. Take 1 L as the basis.
2. Subtract to find the solvent mass.

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

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