JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 36

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Multiple Choice Questions

Question 1: What is the volume at STP of 16 g of ozone (O3)?
(a) 22.4 L
(b) 11.2 L
(c) 7.47 L
(d) 3.73 L
Show Answer & Explanation

Answer: (c) 7.47 L

Explanation:
1. Moles of O3 = 16/48 = 1/3 mol.
2. Volume at STP = (1/3) × 22.4 = 7.47 L.

Teacher's Note:
1. Convert mass to moles with the molar mass, then multiply by 22.4 L.
2. O3 has molar mass 48 g/mol.

Question 2: A gaseous mixture contains 2 mol of CO2, 1 mol of H2 and 2 mol of He. The mole percentage of CO2 is
(a) 40%
(b) 20%
(c) 25%
(d) 50%
Show Answer & Explanation

Answer: (a) 40%

Explanation:
1. Total moles = 2 + 1 + 2 = 5.
2. Mole percent of CO2 = 2/5 × 100 = 40%.

Teacher's Note:
1. Mole fraction = moles of the component / total moles.
2. Multiply by 100 for percent.

Question 3: A compound contains 25% hydrogen and 75% carbon by mass. Its empirical formula is
(a) CH3
(b) C2H5
(c) CH2
(d) CH4
Show Answer & Explanation

Answer: (d) CH4

Explanation:
1. Moles in 100 g: C = 75/12 = 6.25 and H = 25/1 = 25.
2. Ratio C : H = 1 : 4, so the empirical formula is CH4.

Teacher's Note:
1. Divide percentages by atomic masses.
2. Then divide by the smallest number.

Question 4: The mass of iron converted into its oxide by 18 g of steam, 3Fe + 4H2O → Fe3O4 + 4H2 (Fe = 56), is
(a) 42 g
(b) 56 g
(c) 84 g
(d) 21 g
Show Answer & Explanation

Answer: (a) 42 g

Explanation:
1. Moles of steam = 18/18 = 1 mol.
2. From the equation, 4 mol of H2O reacts with 3 mol of Fe, so Fe = 3/4 mol = 42 g.

Teacher's Note:
1. Balance the equation first.
2. Use the mole ratio 3 : 4.

Question 5: 2KClO3 → 2KCl + 3O2. If 448 mL of O2 at NTP is obtained, the mass of KClO3 originally taken is (K = 39, Cl = 35.5, O = 16)
(a) 2.45 g
(b) 3.27 g
(c) 0.82 g
(d) 1.63 g
Show Answer & Explanation

Answer: (d) 1.63 g

Explanation:
1. Moles of O2 = 0.02 mol.
2. Moles of KClO3 = 0.02 × 2/3 = 0.01333 mol, so mass = 0.01333 × 122.5 = 1.63 g.

Teacher's Note:
1. Use the ratio 2 : 3.
2. Molar mass of KClO3 = 122.5 g/mol.

Question 6: 50 g of CaCO3 is allowed to react with 73.5 g of H3PO4: 3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3H2O + 3CO2. The amount of Ca3(PO4)2 formed is
(a) 0.167 mol
(b) 0.5 mol
(c) 0.25 mol
(d) 0.375 mol
Show Answer & Explanation

Answer: (a) 0.167 mol

Explanation:
1. Moles of CaCO3 = 50/100 = 0.5 and moles of H3PO4 = 73.5/98 = 0.75.
2. CaCO3 is limiting (0.5/3 = 0.167 against 0.75/2 = 0.375), so Ca3(PO4)2 = 0.167 mol.

Teacher's Note:
1. Divide the moles by the coefficients to find the limiting reagent.
2. Product follows the limiting reagent.

Question 7: The percent yield of Br2 + Cl2 → 2BrCl in CCl4 is 80%. The moles of BrCl formed from 0.025 mol Br2 and 0.025 mol Cl2 is
(a) 0.040 mol
(b) 0.025 mol
(c) 0.020 mol
(d) 0.050 mol
Show Answer & Explanation

Answer: (a) 0.040 mol

Explanation:
1. The theoretical yield is 2 × 0.025 = 0.050 mol BrCl.
2. Actual yield = 0.80 × 0.050 = 0.040 mol.

Teacher's Note:
1. Actual yield = percent yield x theoretical yield.
2. Each mole of Br2 gives 2 mol of BrCl.

Question 8: KClO3 decomposes by two parallel reactions: (i) 2KClO3 → 2KCl + 3O2, (ii) 4KClO3 → 3KClO4 + KCl. If 3 mol of O2 and 1 mol of KClO4 are produced, the initial moles of KClO3 is
(a) 3 mol
(b) 3.33 mol
(c) 4 mol
(d) 5 mol
Show Answer & Explanation

Answer: (b) 3.33 mol

Explanation:
1. 3 mol of O2 needs 2 mol of KClO3 by reaction (i).
2. 1 mol of KClO4 needs 4/3 mol of KClO3 by reaction (ii), so total = 2 + 4/3 = 3.33 mol.

Teacher's Note:
1. Add the KClO3 consumed in each path.
2. Use the balanced ratios.

Question 9: A 2 g sample of Na2CO3 and NaHCO3 loses 0.248 g when heated, as 2NaHCO3 → Na2CO3 + CO2 + H2O. The mass percentage of Na2CO3 in the mixture is
(a) 87.6%
(b) 33.6%
(c) 12.4%
(d) 66.4%
Show Answer & Explanation

Answer: (d) 66.4%

Explanation:
1. Mass loss of 62 g (CO2 + H2O) accompanies the decomposition of 168 g of NaHCO3.
2. NaHCO3 = 0.248 × 168/62 = 0.672 g, so Na2CO3 = 2 - 0.672 = 1.328 g, which is 66.4%.

Teacher's Note:
1. Only NaHCO3 loses mass.
2. Find its mass from the loss.

Question 10: The mass of KOH needed to prepare 100 mL of 1 M KOH solution is (K = 39)
(a) 0.56 g
(b) 5.6 g
(c) 56 g
(d) 11.2 g
Show Answer & Explanation

Answer: (b) 5.6 g

Explanation:
1. Moles = 1 M x 0.1 L = 0.1 mol.
2. Mass = 0.1 × 56 = 5.6 g.

Teacher's Note:
1. Moles = molarity x volume in litres.
2. Molar mass of KOH = 56 g/mol.

Question 11: A 2 M NaOH solution has density 1 g/mL. Its molality is
(a) 2.17 m
(b) 2.5 m
(c) 1.84 m
(d) 2 m
Show Answer & Explanation

Answer: (a) 2.17 m

Explanation:
1. 1 L of solution weighs 1000 g and contains 2 × 40 = 80 g NaOH.
2. Solvent = 920 g = 0.920 kg, so molality = 2/0.920 = 2.17 m.

Teacher's Note:
1. Work with 1 litre of solution.
2. Subtract solute from the solution mass.

Question 12: The mole fraction of NaOH in a 2 M NaOH solution of density 1 g/mL is
(a) 0.0826
(b) 0.0357
(c) 0.0200
(d) 0.0377
Show Answer & Explanation

Answer: (d) 0.0377

Explanation:
1. 1 L of solution contains 2 mol NaOH and 920 g water = 51.1 mol.
2. x = 2/(2 + 51.1) = 0.0377.

Teacher's Note:
1. Count moles of water from the solvent mass.
2. x = n(solute) / total moles.

Question 13: The mass percentage (w/w) of NaOH in a 2 M solution of density 1 g/mL is
(a) 4%
(b) 20%
(c) 16%
(d) 8%
Show Answer & Explanation

Answer: (d) 8%

Explanation:
1. 1 L of solution has mass 1000 g and contains 80 g NaOH.
2. w/w = 80/1000 × 100 = 8%.

Teacher's Note:
1. Use 1 litre as the basis.
2. Mass of solution = volume x density.

Question 14: For a 5 molal NaOH solution of density 1.5 g/mL, the % (w/v) of NaOH is
(a) 16.7%
(b) 12.5%
(c) 20%
(d) 25%
Show Answer & Explanation

Answer: (d) 25%

Explanation:
1. 200 g NaOH is in 1200 g of solution, whose volume is 800 mL.
2. w/v = 200/800 × 100 = 25%.

Teacher's Note:
1. w/v needs the volume of solution.
2. Volume = 1200/1.5 = 800 mL.

Question 15: The volume of water to be added to 50 mL of HNO3 (density 1.5 g/mL, 63% by weight) to make a 1 M solution is (assume volumes are additive)
(a) 750 mL
(b) 1000 mL
(c) 650 mL
(d) 700 mL
Show Answer & Explanation

Answer: (d) 700 mL

Explanation:
1. Mass of acid solution = 75 g, so HNO3 = 0.63 × 75 = 47.25 g = 0.75 mol.
2. For 1 M the total volume = 0.75 L = 750 mL, so water added = 750 - 50 = 700 mL.

Teacher's Note:
1. Moles of solute stay the same on dilution.
2. Total volume = moles / molarity.

Question 16: 300 g of 30% (w/w) NaOH is mixed with 500 g of 40% (w/w) NaOH. The mass percentage of NaOH in the final solution is
(a) 36.25%
(b) 35%
(c) 38%
(d) 34.5%
Show Answer & Explanation

Answer: (a) 36.25%

Explanation:
1. NaOH = 0.30 × 300 + 0.40 × 500 = 90 + 200 = 290 g in 800 g.
2. % = 290/800 × 100 = 36.25%.

Teacher's Note:
1. Add the masses of solute and of solution.
2. Then take the ratio.

Question 17: For the solution obtained by mixing 300 g of 30% and 500 g of 40% (w/w) NaOH, the molality is
(a) 14.2 m
(b) 7.25 m
(c) 9.06 m
(d) 18.1 m
Show Answer & Explanation

Answer: (a) 14.2 m

Explanation:
1. NaOH = 290 g = 7.25 mol and solvent = 800 - 290 = 510 g.
2. Molality = 7.25/0.510 = 14.2 m.

Teacher's Note:
1. Molality uses the mass of the solvent in kg.
2. Moles = 290/40.

Question 18: In acidic solution HSO4- + As4 + Pb3O4 → PbSO4 + H2AsO4-. In the balanced equation the coefficient of Pb3O4 per As4 is
(a) 4
(b) 5
(c) 10
(d) 20
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. One As atom changes from 0 to +5, so As4 loses 20 e-.
2. Pb3O4 (two Pb(II) and one Pb(IV)) gains 2 e-, so 10 Pb3O4 are needed per As4.
3. 10Pb3O4 + As4 + 30HSO4- + 26H+ → 30PbSO4 + 4H2AsO4- + 24H2O.

Teacher's Note:
1. Count electrons lost and gained.
2. The ratio of the two gives the coefficients.

Question 19: In basic solution H2O2 + Cl2O7 → ClO2- + O2. In the balanced equation the coefficients of H2O2, O2 and H2O are respectively
(a) 2, 2 and 3
(b) 3, 4 and 4
(c) 4, 4 and 5
(d) 4, 2 and 5
Show Answer & Explanation

Answer: (c) 4, 4 and 5

Explanation:
1. Each Cl goes from +7 to +3, so Cl2O7 gains 8 e-; H2O2 loses 2 e- per O2, so 4 H2O2 are needed.
2. 4H2O2 + Cl2O7 + 2OH- → 2ClO2- + 4O2 + 5H2O.

Teacher's Note:
1. Hydrogen peroxide acts as a reductant here.
2. Balance O and H with water and OH-.

Question 20: The analysis of three binary oxides of bromine gives masses of oxygen combined with 1.0 g of Br of 0.101 g (X), 0.303 g (Y) and 0.503 g (Z). Which of the following statements is not correct? (I) Y is Br2O3, (II) Z is Br2O5, (III) Z is Br2O7, (IV) Y is Br2O5
(a) III and IV
(b) II and IV
(c) I and III
(d) I and II
Show Answer & Explanation

Answer: (a) III and IV

Explanation:
1. Moles of Br = 1/79.9 = 0.01252; moles of O = 0.00631, 0.01894 and 0.03144 for X, Y and Z.
2. O : Br = 0.50, 1.51 and 2.51, so X = Br2O, Y = Br2O3 and Z = Br2O5; hence III and IV are incorrect.

Teacher's Note:
1. Convert the masses to mole ratios.
2. Multiply by 2 to get whole numbers.

Question 21: Myoglobin contains 0.34% Fe by mass and each molecule has one Fe ion (molar mass of Fe = 55.845 g/mol). The molar mass of myoglobin (g/mol) is
(a) 164206
(b) 164250
(c) 16407
(d) 16425
Show Answer & Explanation

Answer: (d) 16425

Explanation:
1. M = 55.845/0.0034 = 16 425 g/mol.

Teacher's Note:
1. Molar mass = atomic mass / mass fraction.
2. One Fe per molecule.

Question 22: The permissible level of mercury in drinking water is 0.0335 ppb (1 ppb = 1 microgram per litre). An equivalent representation is
(a) 3.35 × 10-5 mg dm-3
(b) 3.35 × 10-2 mg dm-3
(c) 3.35 × 10-5 mg m-3
(d) 3.35 × 10-4 g L-1
Show Answer & Explanation

Answer: (a) 3.35 × 10-5 mg dm-3

Explanation:
1. 0.0335 microgram/L = 3.35 × 10-2 microgram/L = 3.35 × 10-5 mg/L, and 1 L = 1 dm3.
2. It equals 3.35 × 10-2 mg m-3, so the other options are wrong.

Teacher's Note:
1. 1 microgram = 10-3 mg.
2. 1 m3 = 1000 L.

Question 23: Lipoic acid is 1,2-dithiolane-3-pentanoic acid, C8H14O2S2. Taking C = 12, H = 1, S = 32.065 and O = 15.999, the percentages of S and O are respectively
(a) 33.03 and 16.48
(b) 31.11 and 18.24
(c) 31.42 and 15.68
(d) 31.11 and 15.52
Show Answer & Explanation

Answer: (d) 31.11 and 15.52

Explanation:
1. M = 8(12) + 14(1) + 2(15.999) + 2(32.065) = 206.13 g/mol.
2. %S = 64.13/206.13 = 31.11% and %O = 32.00/206.13 = 15.52%.

Teacher's Note:
1. Find the formula from the structure: 8 C, 14 H, 2 O, 2 S.
2. Mass percent = part / whole x 100.

Question 24: The number of water molecules in a 0.20 g sample of CuSO4.5H2O (molar mass 249.7 g/mol) is
(a) 1.2 × 1021
(b) 2.41 × 1022
(c) 1.2 × 1023
(d) 2.41 × 1021
Show Answer & Explanation

Answer: (d) 2.41 × 1021

Explanation:
1. Moles of the hydrate = 0.20/249.7 = 8.01 × 10-4.
2. Water molecules = 5 × 8.01 × 10-4 x 6.022 × 1023 = 2.41 × 1021.

Teacher's Note:
1. Each formula unit has five water molecules.
2. Use NA = 6.022 × 1023.

Question 25: The measured density of helium at NTP is 0.1784 g/L. The mass of one mole of helium (in g) is
(a) 2.0 g
(b) 8.0 g
(c) 4.0 g
(d) 0.1784 g
Show Answer & Explanation

Answer: (c) 4.0 g

Explanation:
1. Mass of 22.4 L = 0.1784 × 22.4 = 3.996 g, which is about 4.0 g.

Teacher's Note:
1. One mole occupies 22.4 L at NTP.
2. Helium is monoatomic with M = 4.

Question 26: In I2 + C2H5OH + OH- → CHI3 + HCOO- + H2O + I- (one CHI3 and one HCOO- per ethanol), which statements are correct? (A) the coefficients of OH- and I- are 6 and 5; (B) they are 5 and 6; (C) ethanol is oxidised to CHI3 and HCOO-; (D) 8 electrons are involved in the conversion of ethanol
(a) B, C and D only
(b) A and D only
(c) A, C and D only
(d) A and C only
Show Answer & Explanation

Answer: (c) A, C and D only

Explanation:
1. Balanced: 4I2 + C2H5OH + 6OH- → CHI3 + HCOO- + 5I- + 5H2O.
2. Carbon in ethanol (average -2 each, total -4) goes to +2 and +2 (total +4): 8 e- are lost, which 4 I2 accept.

Teacher's Note:
1. Balance with electrons: 4 I2 gain 8 e-.
2. Check atoms and charge.

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