JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 04

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: The total number of ions present in 1 mL of 0.1 M barium nitrate, Ba(NO3)2, solution is
(a) 6.02 × 1018
(b) 6.02 × 1019
(c) 3.0 × 6.02 × 1019
(d) 3.0 × 6.02 × 1018
Show Answer & Explanation

Answer: (c) 3.0 × 6.02 × 1019

Explanation:
1. Moles of Ba(NO3)2 = 0.1 × 1/1000 = 10-4 mol.
2. Each formula unit gives 3 ions (1 Ba2+ and 2 NO3-).
3. Ions = 3 × 10-4 × 6.02 × 1023 = 3.0 × 6.02 × 1019.

Teacher's Note:
1. Molarity × volume in litres gives moles.
2. Count all ions produced per formula unit.

Question 2: The weight of 1 mole of a gas of density 0.1784 g L-1 at NTP is
(a) 0.1784 g
(b) 1 g
(c) 4 g
(d) 4 amu
Show Answer & Explanation

Answer: (c) 4 g

Explanation:
1. Mass of 1 mol = density × molar volume = 0.1784 × 22.4 = 4.0 g.

Teacher's Note:
1. Molar mass = density × 22.4 at STP.
2. This is helium, with molar mass 4 g/mol.

Question 3: Given that one mole of N2 at NTP occupies 22.4 L, the density of N2 is
(a) 1.25 g L-1
(b) 0.80 g L-1
(c) 2.5 g L-1
(d) 1.60 g L-1
Show Answer & Explanation

Answer: (a) 1.25 g L-1

Explanation:
1. Density = 28 g / 22.4 L = 1.25 g/L.

Teacher's Note:
1. Density of a gas at STP = molar mass ÷ 22.4.
2. Nitrogen is slightly lighter than air.

Question 4: The number of gram molecules of oxygen in 6.02 × 1024 CO molecules is
(a) 10 g molecules
(b) 5 g molecules
(c) 1 g molecule
(d) 0.5 g molecule
Show Answer & Explanation

Answer: (b) 5 g molecules

Explanation:
1. Moles of CO = 6.02 × 1024/6.02 × 1023 = 10 mol.
2. Each CO has one O atom, so there are 10 mol of O atoms.
3. One gram molecule of O2 holds 2 mol of atoms, so 10 mol of atoms = 5 gram molecules.

Teacher's Note:
1. A gram molecule is one mole of molecules.
2. Convert atoms of oxygen into O2 molecules.

Question 5: A compound of X and Y has equal masses of the two elements. If their atomic weights are 30 and 20 respectively, the molecular formula of the compound is
(a) X2Y2
(b) X3Y3
(c) X2Y3
(d) X3Y2
Show Answer & Explanation

Answer: (c) X2Y3

Explanation:
1. Take 60 g of each: moles X = 60/30 = 2, moles Y = 60/20 = 3.
2. Mole ratio X : Y = 2 : 3, so the formula is X2Y3.

Teacher's Note:
1. Divide each mass by atomic mass and take the simplest ratio.
2. The formula follows the mole ratio.

Question 6: An oxide of sulphur contains 50% of sulphur. Its empirical formula is
(a) SO2
(b) SO3
(c) SO
(d) S2O
Show Answer & Explanation

Answer: (a) SO2

Explanation:
1. In 100 g: S = 50/32 = 1.5625 mol, O = 50/16 = 3.125 mol.
2. Ratio S : O = 1 : 2, so the empirical formula is SO2.

Teacher's Note:
1. Percentage composition converts to moles by dividing with atomic mass.
2. Divide by the smallest to get the ratio.

Question 7: A hydrocarbon contains 80% carbon. The hydrocarbon is
(a) CH4
(b) C2H4
(c) C2H6
(d) C2H2
Show Answer & Explanation

Answer: (c) C2H6

Explanation:
1. For C2H6: 24/30 × 100 = 80%.
2. CH4 has 75%, C2H4 has 85.7% and C2H2 has 92.3%.

Teacher's Note:
1. Calculate the percentage of carbon for each option.
2. C2H6 is the only match.

Question 8: The empirical formula of glucose is
(a) C6H12O6
(b) C3H6O3
(c) C2H4O2
(d) CH2O
Show Answer & Explanation

Answer: (d) CH2O

Explanation:
1. Glucose is C6H12O6, with ratio C : H : O = 6 : 12 : 6 = 1 : 2 : 1.
2. The empirical formula is CH2O.

Teacher's Note:
1. Divide all subscripts by their highest common factor.
2. C2H4O2 and C3H6O3 also have CH2O as their empirical formula.

Question 9: An oxide of metal M has 40% oxygen by mass. Metal M has atomic mass 24. The empirical formula of the oxide is
(a) M2O
(b) M2O3
(c) MO
(d) M3O4
Show Answer & Explanation

Answer: (c) MO

Explanation:
1. In 100 g: M = 60/24 = 2.5 mol and O = 40/16 = 2.5 mol.
2. The ratio is 1 : 1, so the formula is MO.

Teacher's Note:
1. Use the metal's percentage as 100 minus the oxygen percentage.
2. Equal moles give a 1 : 1 formula.

Question 10: A compound contains 38.8% C, 16.0% H and 45.2% N. The formula of the compound would be
(a) CH3NH2
(b) CH3CN
(c) C2H5CN
(d) CH2(NH)2
Show Answer & Explanation

Answer: (a) CH3NH2

Explanation:
1. Moles: C = 38.8/12 = 3.23, H = 16.0/1 = 16.0, N = 45.2/14 = 3.23.
2. Ratio C : H : N = 1 : 5 : 1, which is CH5N, i.e. CH3NH2.

Teacher's Note:
1. Write the empirical ratio first, then match the options.
2. Option 1 has 5 H atoms in total.

Question 11: The simplest formula of a compound containing 50% of element X (atomic weight 10) and 50% of element Y (atomic weight 20) is
(a) XY
(b) X2Y
(c) XY2
(d) X3Y
Show Answer & Explanation

Answer: (b) X2Y

Explanation:
1. Moles X = 50/10 = 5 and moles Y = 50/20 = 2.5.
2. Ratio X : Y = 2 : 1, so the formula is X2Y.

Teacher's Note:
1. Divide the percentage by the atomic weight.
2. Then divide by the smaller number.

Question 12: Which of the following compounds has the same empirical formula as that of glucose?
(a) CH3CHO
(b) CH3COOH
(c) CH3OH
(d) C2H6
Show Answer & Explanation

Answer: (b) CH3COOH

Explanation:
1. The empirical formula of glucose is CH2O.
2. CH3COOH (C2H4O2) reduces to CH2O.

Teacher's Note:
1. Reduce each option to its simplest ratio.
2. Acetic acid and glucose share the same empirical formula.

Question 13: A gas is found to contain 2.34 g of nitrogen and 5.34 g of oxygen. The simplest formula of the compound is
(a) N2O
(b) NO
(c) N2O3
(d) NO2
Show Answer & Explanation

Answer: (d) NO2

Explanation:
1. Moles N = 2.34/14 = 0.167 and moles O = 5.34/16 = 0.334.
2. Ratio N : O = 1 : 2, so the formula is NO2.

Teacher's Note:
1. Convert masses to moles and divide by the smaller value.
2. NO2 is a brown gas.

Question 14: 2.2 g of a compound of phosphorus and sulphur has 1.24 g of P in it. Its empirical formula is
(a) P2S3
(b) P3S2
(c) P3S4
(d) P4S3
Show Answer & Explanation

Answer: (d) P4S3

Explanation:
1. Mass of S = 2.2 − 1.24 = 0.96 g.
2. Moles P = 1.24/31 = 0.04 and moles S = 0.96/32 = 0.03.
3. Ratio P : S = 4 : 3, so the formula is P4S3.

Teacher's Note:
1. Find the second element by subtraction.
2. P4S3 is a known sulphide of phosphorus.

Question 15: On analysis, a certain compound was found to contain iodine and oxygen in the ratio 254 : 80. The formula of the compound is (At. mass: I = 127, O = 16)
(a) IO
(b) I2O
(c) I5O2
(d) I2O5
Show Answer & Explanation

Answer: (d) I2O5

Explanation:
1. Moles I = 254/127 = 2 and moles O = 80/16 = 5.
2. Ratio I : O = 2 : 5, so the formula is I2O5.

Teacher's Note:
1. The ratio of masses is used like percentages.
2. Iodine pentoxide is I2O5.

Question 16: The numbers of atoms of Cr and O are 4.8 × 1010 and 9.6 × 1010 respectively. Its empirical formula is
(a) Cr2O3
(b) CrO2
(c) Cr2O4
(d) CrO5
Show Answer & Explanation

Answer: (b) CrO2

Explanation:
1. Ratio Cr : O = 4.8 : 9.6 = 1 : 2.
2. The empirical formula is CrO2.

Teacher's Note:
1. When atom counts are given, divide them directly.
2. Cr2O4 is not in simplest form.

Question 17: Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is
(a) 941.176
(b) 944
(c) 945.27
(d) None of these
Show Answer & Explanation

Answer: (a) 941.176

Explanation:
1. One S atom (32 g) is 3.4% of the molecule, so minimum mass = 32 × 100/3.4 = 941.18.

Teacher's Note:
1. Minimum molecular mass assumes one atom of the given element.
2. Mass = atomic mass × 100 ÷ percentage.

Question 18: A giant molecule contains 0.25% of a metal whose atomic weight is 59. Its molecule contains one atom of that metal. Its minimum molecular weight is
(a) 5900
(b) 23600
(c) 11800
(d) \( \frac{100 \times 59}{0.4} \)
Show Answer & Explanation

Answer: (b) 23600

Explanation:
1. Molecular mass = 59 × 100/0.25 = 23600.

Teacher's Note:
1. Use the formula atomic mass × 100 ÷ percentage.
2. 0.25 divides 100 exactly 400 times.

Question 19: Caffeine has a molecular weight of 194. It contains 28.9% by mass of nitrogen. The number of atoms of nitrogen in one molecule of it is
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Mass of N in one molecule = 0.289 × 194 = 56 g.
2. Number of N atoms = 56/14 = 4.

Teacher's Note:
1. Mass of the element in one mole divided by atomic mass gives the atom count.
2. Round to the nearest whole number.

Question 20: In a gaseous reaction of the type aA + bB → cC + dD, which statement is wrong?
(a) a litre of A combines with b litre of B to give C and D
(b) a mole of A combines with b moles of B to give C and D
(c) a g of A combines with b g of B to give C and D
(d) a molecules of A combine with b molecules of B to give C and D
Show Answer & Explanation

Answer: (c) a g of A combines with b g of B to give C and D

Explanation:
1. The coefficients show the ratio of moles, molecules or volumes (for gases), not of grams.
2. So the statement about a g and b g is wrong.

Teacher's Note:
1. Coefficients are mole ratios, never mass ratios.
2. For gases at the same T and P they are also volume ratios.

Question 21: Assuming that petrol is octane (C8H18) with density 0.8 g mL-1, 1.425 L of petrol on complete combustion will consume
(a) 50 mol of O2
(b) 100 mol of O2
(c) 125 mol of O2
(d) 200 mol of O2
Show Answer & Explanation

Answer: (c) 125 mol of O2

Explanation:
1. Mass of petrol = 1425 mL × 0.8 = 1140 g, i.e. 1140/114 = 10 mol.
2. C8H18 + 12.5 O2 → 8CO2 + 9H2O, so O2 = 10 × 12.5 = 125 mol.

Teacher's Note:
1. Balance the combustion equation first.
2. Convert volume to mass with the density.

Question 22: In the reaction 2Al + 3/2 O2 → Al2O3, 9 g of Al will react with
(a) 6 g O2
(b) 8 g O2
(c) 9 g O2
(d) 4 g O2
Show Answer & Explanation

Answer: (b) 8 g O2

Explanation:
1. Moles Al = 9/27 = 1/3 mol.
2. O2 needed = (3/4) × (1/3) = 1/4 mol = 8 g.

Teacher's Note:
1. Use the mole ratio Al : O2 = 2 : 1.5.
2. 1/4 mol of O2 weighs 8 g.

Question 23: The equation 2Al(s) + 3/2 O2(g) → Al2O3(s) shows that
(a) 2 mol of Al reacts with 3/2 mol of O2 to produce 7/2 mol of Al2O3
(b) 2 g of Al reacts with 3/2 g of O2 to produce one mol of Al2O3
(c) 2 g of Al reacts with 3/2 L of O2 to produce 1 mol of Al2O3
(d) 2 mol of Al reacts with 3/2 mol of O2 to produce 1 mol of Al2O3
Show Answer & Explanation

Answer: (d) 2 mol of Al reacts with 3/2 mol of O2 to produce 1 mol of Al2O3

Explanation:
1. The coefficients are mole ratios: 2 mol Al + 1.5 mol O2 gives 1 mol Al2O3.
2. The other options mix up grams, litres and moles.

Teacher's Note:
1. Read the equation in moles unless told otherwise.
2. Check that mass is conserved on both sides.

Question 24: 1 L of CO2 is passed over hot coke. When the volume of the reaction mixture becomes 1.4 L, the composition of the reaction mixture is
(a) 0.6 L CO
(b) 0.8 L CO2
(c) 0.6 L CO2 and 0.8 L CO
(d) None of these
Show Answer & Explanation

Answer: (c) 0.6 L CO2 and 0.8 L CO

Explanation:
1. CO2 + C → 2CO: if x L of CO2 reacts, the volume rises by x.
2. 1 + x = 1.4, so x = 0.4 L; CO formed = 0.8 L and CO2 left = 0.6 L.

Teacher's Note:
1. Volume increase equals the volume of CO2 consumed.
2. Write the final mixture as CO2 left plus CO formed.

Question 25: 26 cc of CO2 are passed over red hot coke. The volume of CO evolved is
(a) 15 cc
(b) 10 cc
(c) 32 cc
(d) 52 cc
Show Answer & Explanation

Answer: (d) 52 cc

Explanation:
1. CO2 + C → 2CO, so 1 volume of CO2 gives 2 volumes of CO.
2. CO = 2 × 26 = 52 cc.

Teacher's Note:
1. Gas volumes follow the mole ratio at the same T and P.
2. Assume complete conversion unless told otherwise.

Question 26: If 1/2 mol of oxygen combines with aluminium to form Al2O3, the weight of aluminium metal used in the reaction is (Al = 27)
(a) 27 g
(b) 18 g
(c) 54 g
(d) 40.5 g
Show Answer & Explanation

Answer: (b) 18 g

Explanation:
1. 4Al + 3O2 → 2Al2O3, so 3 mol O2 needs 4 mol Al.
2. Al = (4/3) × 0.5 = 2/3 mol = (2/3) × 27 = 18 g.

Teacher's Note:
1. Use the balanced equation 4Al + 3O2.
2. Mole ratio Al : O2 = 4 : 3.

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