JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 35

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: C2H5OH + C2H5COOH → C2H5COOC2H5 + H2O. 349 g of ethyl propanoate (102 g mol-1) was obtained from 250 g of ethanol (46 g mol-1) with excess acid. The percentage yield is
(a) 48.2
(b) 32.7
(c) 62.9
(d) 54.6
Show Answer & Explanation

Answer: (c) 62.9

Explanation:
1. Ethanol = 250/46 = 5.43 mol, so theoretical ester = 5.43 × 102 = 554 g.
2. Yield = 349/554 × 100 = 62.9%.

Teacher's Note:
1. Yield = actual / theoretical.

Question 2: Which of the following has the maximum percentage of sulphur by mass?
(a) Na2SO4
(b) Li2SO4
(c) PbSO4
(d) H2SO4
Show Answer & Explanation

Answer: (d) H2SO4

Explanation:
1. % S: Na2SO4 22.5; H2SO4 32.7; Li2SO4 29.1; PbSO4 10.6.
2. H2SO4 has the highest percentage.

Teacher's Note:
1. The lighter the rest of the formula, the larger the S fraction.

Question 3: MnO2 + 4HCl → Cl2 + MnCl2 + 2H2O. The mass of concentrated HCl solution (36.0% HCl by mass) needed to produce 2.50 g of Cl2 is
(a) 26.4 g
(b) 14.3 g
(c) 5.15 g
(d) 19.4 g
Show Answer & Explanation

Answer: (b) 14.3 g

Explanation:
1. Cl2 = 2.50/71 = 0.0352 mol, needing 4 × 0.0352 = 0.141 mol HCl = 5.14 g.
2. Solution mass = 5.14/0.36 = 14.3 g.

Teacher's Note:
1. Divide the mass of HCl by the mass fraction.

Question 4: How many moles of Na+ ions are there in 20 mL of 0.40 M Na3PO4?
(a) 0.008
(b) 0.024
(c) 0.040
(d) 0.020
Show Answer & Explanation

Answer: (b) 0.024

Explanation:
1. Moles of Na3PO4 = 0.020 × 0.40 = 0.008 mol.
2. Moles of Na+ = 3 × 0.008 = 0.024 mol.

Teacher's Note:
1. Each formula unit has three sodium ions.

Question 5: The Na+ concentration in the solution formed by mixing 20 mL of 0.10 M Na2SO4 with 50 mL of 0.30 M Na3PO4 is
(a) 0.70 M
(b) 0.48 M
(c) 0.24 M
(d) 0.15 M
Show Answer & Explanation

Answer: (a) 0.70 M

Explanation:
1. Na+ = 20 × 0.10 × 2 + 50 × 0.30 × 3 = 4 + 45 = 49 mmol.
2. Total volume = 70 mL, so concentration = 49/70 = 0.70 M.

Teacher's Note:
1. Add the moles of sodium from both salts.

Question 6: The number of moles of oxygen needed to burn 1.0 mol of methyl tert-butyl ether, C5H12O, completely to CO2 and H2O is
(a) 7.5 mol
(b) 6.0 mol
(c) 8.0 mol
(d) 4.5 mol
Show Answer & Explanation

Answer: (a) 7.5 mol

Explanation:
1. C5H12O + 15/2 O2 → 5CO2 + 6H2O.
2. O needed on the right = 10 + 6 = 16 atoms; the compound supplies 1, so 15 atoms = 7.5 mol O2.

Teacher's Note:
1. Count the oxygen atoms on both sides.

Question 7: The hydrated salt Na2SO4·nH2O loses all its water on heating and is reduced to 44.1% of its original mass. The value of n is
(a) 5
(b) 7
(c) 10
(d) 6
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. 142/(142 + 18n) = 0.441, so 142 + 18n = 322 and n = 10.

Teacher's Note:
1. Anhydrous mass fraction = M(anhydrous) / M(hydrate).

Question 8: The simplest formula of a compound containing 50% of element A (atomic mass 10) and 50% of element B (atomic mass 20) is
(a) A2B
(b) A2B3
(c) A2B2
(d) AB
Show Answer & Explanation

Answer: (a) A2B

Explanation:
1. Moles in 100 g: A = 5, B = 2.5.
2. Ratio A : B = 2 : 1, so the formula is A2B.

Teacher's Note:
1. Divide by the smaller number of moles.

Question 9: Heating a solution does not change
(a) the molarity of the solution
(b) the molality of the solution
(c) the density of the solution
(d) the normality of the solution
Show Answer & Explanation

Answer: (b) the molality of the solution

Explanation:
1. Heating changes the volume, so molarity, normality and density change.
2. The masses of solute and solvent do not change, so molality is unchanged.

Teacher's Note:
1. Mass-based terms are temperature independent.

Question 10: 0.14 g of a substance burnt in oxygen yields 0.28 g of oxide. The substance is
(a) carbon
(b) nitrogen
(c) sulphur
(d) phosphorus
Show Answer & Explanation

Answer: (c) sulphur

Explanation:
1. Oxygen combined = 0.14 g = 0.00875 mol of O atoms.
2. For sulphur: 0.14/32 = 0.004375 mol, ratio O : S = 2, which gives SO2.
3. For N, C and P, the ratio of O atoms to element atoms is 0.88, 0.75 and 1.94, which are not a simple whole-number ratio of a known oxide.

Teacher's Note:
1. Test each element for a simple whole-number ratio.

Question 11: The number of molecules of water of hydration in 252 mg of oxalic acid dihydrate, H2C2O4·2H2O (M = 126 g mol-1), is
(a) 2.52 × 1021
(b) 2.68 × 1018
(c) 1.83 × 1024
(d) 2.4 × 1021
Show Answer & Explanation

Answer: (d) 2.4 × 1021

Explanation:
1. Moles of hydrate = 0.252/126 = 2.0 × 10-3 mol.
2. Water molecules = 2 × 2.0 × 10-3 × 6.02 × 1023 = 2.4 × 1021.

Teacher's Note:
1. Two water molecules per formula unit.

Question 12: A compound of beryllium has Be 6.1%, N 37.8%, Cl 48%, H 8.1%. One mole has mass 148 g and the average atomic mass of Be is 9. The molecular formula is
(a) BeN2H10Cl
(b) BeN4H12Cl2
(c) Be2N4H10Cl2
(d) BeN4H2Cl3
Show Answer & Explanation

Answer: (b) BeN4H12Cl2

Explanation:
1. Atoms per mole: Be = 148 × 0.061/9 = 1.0; N = 148 × 0.378/14 = 4.0; Cl = 148 × 0.48/35.5 = 2.0; H = 148 × 0.081/1 = 12.
2. Formula = BeN4H12Cl2.

Teacher's Note:
1. Divide the mass of each element in 148 g by its atomic mass.

Question 13: The molarity of 20% w/w sulphuric acid of density 1.14 g cm-3 is
(a) 2.22
(b) 2.32
(c) 2.02
(d) 2.12
Show Answer & Explanation

Answer: (b) 2.32

Explanation:
1. M = 10 × 1.14 × 20/98 = 2.33, closest to 2.32.

Teacher's Note:
1. Use M = 10 d (%)/M_2.

Question 14: 2.00 g of a sample containing a bromide impurity required 6.4 mL of 0.20 M AgNO3 to precipitate the bromide completely. The mass percentage of the impurity as bromide is (Br = 80)
(a) 2.56
(b) 5.11
(c) 1.28
(d) 9.15
Show Answer & Explanation

Answer: (b) 5.11

Explanation:
1. AgNO3 = 6.4 × 0.20 = 1.28 mmol = moles of Br-.
2. Mass of bromide = 1.28 × 10-3 × 80 = 0.1024 g.
3. Percentage = 0.1024/2.00 × 100 = 5.1%.

Teacher's Note:
1. Silver ions react 1 : 1 with bromide.

Question 15: The largest number of molecules is present in
(a) 64 g of N2O5
(b) 36 g of H2O
(c) 70 g of SO2
(d) 34 g of CO2
Show Answer & Explanation

Answer: (b) 36 g of H2O

Explanation:
1. Moles: SO2 = 1.09; N2O5 = 64/108 = 0.59; H2O = 2.0; CO2 = 0.77.
2. H2O has the most.

Teacher's Note:
1. Divide each mass by the molar mass.

Question 16: An element X combines with oxygen to form X4O6. If 8.40 g of X combines with 6.50 g of oxygen, the atomic mass of X (u) is
(a) 24.0
(b) 50.4
(c) 31.0
(d) 118.7
Show Answer & Explanation

Answer: (c) 31.0

Explanation:
1. O atoms = 6.50/16 = 0.406 mol, so X atoms = 0.406 × 4/6 = 0.271 mol.
2. Atomic mass = 8.40/0.271 = 31.0 u (phosphorus).

Teacher's Note:
1. Use the ratio X : O = 4 : 6.

Question 17: Excess AgNO3 is added to a water sample and 1.4 g of AgCl (143.25 g mol-1) is precipitated. The mass of chloride ion in the sample is
(a) 0.25 g
(b) 0.75 g
(c) 0.35 g
(d) 0.50 g
Show Answer & Explanation

Answer: (c) 0.35 g

Explanation:
1. Moles of AgCl = 1.4/143.25 = 9.77 × 10-3 mol = moles of Cl-.
2. Mass = 9.77 × 10-3 × 35.5 = 0.347 g, about 0.35 g.

Teacher's Note:
1. Each AgCl contains one chloride ion.

Question 18: The mode of expressing concentration that is independent of temperature is
(a) Normality
(b) Molality
(c) Formality
(d) Molarity
Show Answer & Explanation

Answer: (b) Molality

Explanation:
1. Molarity, normality and formality are based on volume, which varies with temperature.
2. Molality is based on masses only.

Teacher's Note:
1. Mass-based terms are independent of temperature.

Question 19: In a nitration experiment, 10.0 g of benzene gave 13.2 g of nitrobenzene (C6H6 → C6H5NO2; 78 and 123 g mol-1). The percentage yield is
(a) 62.7%
(b) 26.7%
(c) 83.5%
(d) 88.9%
Show Answer & Explanation

Answer: (c) 83.5%

Explanation:
1. Benzene = 10/78 = 0.128 mol, so theoretical nitrobenzene = 0.128 × 123 = 15.8 g.
2. Yield = 13.2/15.8 × 100 = 83.7%, closest to 83.5%.

Teacher's Note:
1. Percentage yield = actual / theoretical.

Question 20: The number of moles of hydrogen atoms in 1.006 × 1023 molecules of diethyl ether is about
(a) 0.16
(b) 3
(c) 1.67
(d) 6
Show Answer & Explanation

Answer: (c) 1.67

Explanation:
1. Moles of ether = 1.006 × 1023/6.022 × 1023 = 0.167 mol.
2. Diethyl ether, C4H10O, has 10 H atoms, so H = 1.67 mol.

Teacher's Note:
1. Atoms per molecule times moles of molecules.

Question 21: The specific gravity of a HNO3 solution is 1.42 and it is 70% w/w. The molarity is
(a) 15.8
(b) 11.1
(c) 31.6
(d) 14.2
Show Answer & Explanation

Answer: (a) 15.8

Explanation:
1. M = 10 × 1.42 × 70/63 = 15.8 M.

Teacher's Note:
1. Use M = 10 d (%)/M_2.

Question 22: The ratio of the masses of methane and ethane in a gas mixture is 4 : 5. The ratio of the numbers of their molecules is
(a) 2 : 3
(b) 4 : 5
(c) 5 : 4
(d) 3 : 2
Show Answer & Explanation

Answer: (d) 3 : 2

Explanation:
1. Moles: methane = 4/16 = 0.25; ethane = 5/30 = 0.1667.
2. Ratio = 0.25 : 0.1667 = 3 : 2.

Teacher's Note:
1. Number of molecules is proportional to moles.

Question 23: At constant T and P, 5.0 L of SO2 reacts with 3.0 L of O2: 2SO2 + O2 → 2SO3. The volume of the mixture on completion of the reaction is
(a) 5 L
(b) 8.0 L
(c) 0.5 L
(d) 5.5 L
Show Answer & Explanation

Answer: (d) 5.5 L

Explanation:
1. 5.0 L SO2 needs 2.5 L O2 and gives 5.0 L SO3.
2. O2 left = 3.0 - 2.5 = 0.5 L, so total volume = 5.0 + 0.5 = 5.5 L.

Teacher's Note:
1. Gas volumes at the same T and P are in the ratio of moles.

Question 24: Which mixture of water and H2SO4 has a mass percentage of H2SO4 close to 30?
(a) 30 g H2SO4 + 100 g H2O
(b) 1 mol H2SO4 + 2 mol H2O
(c) 1 mol H2SO4 + 200 g H2O
(d) 0.30 mol H2SO4 + 0.70 mol H2O
Show Answer & Explanation

Answer: (c) 1 mol H2SO4 + 200 g H2O

Explanation:
1. (a) 30/130 = 23%. (b) 98/(98 + 36) = 73%.
2. (c) 98/(98 + 200) = 32.9%. (d) 29.4/(29.4 + 12.6) = 70%.
3. Option (c) is closest to 30%.

Teacher's Note:
1. Convert all moles to masses.

Question 25: (CH3)2NNH2 and N2O4 react stoichiometrically to give only N2, CO2 and H2O (all gases). The number of moles of gas produced from 1 mol of (CH3)2NNH2 is
(a) 3
(b) 6
(c) 4.5
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. (CH3)2NNH2 + 2N2O4 → 2CO2 + 4H2O + 3N2.
2. Gas moles = 2 + 4 + 3 = 9.

Teacher's Note:
1. Balance C, H, N and O in turn.

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