JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 21

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Chapter-wise Objective Questions: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

Question 1: A chemist finds a compound with H : O atoms in the ratio 2 : 1, 40% C by mass, approximate molecular mass 178 g, and containing only C, H and O. The percentage by mass of oxygen in the compound is
(a) 53.33%
(b) 88.88%
(c) 33.33%
(d) 40.00%
Show Answer & Explanation

Answer: (a) 53.33%

Explanation:
1. Let O = y atoms-worth: H = 2y. Mass: H + O = 60% so 2y + 16y = 60.
2. y = 3.33, so O = 53.33%.

Teacher's Note:
1. The remaining 60% is shared by hydrogen and oxygen in an atom ratio 2 : 1.

Question 2: A chemist finds a compound with H : O atoms in the ratio 2 : 1, 40% C by mass, approximate molecular mass 178 g, and containing only C, H and O. Which of the following could be the molecular formula of the compound?
(a) \( \ce{C6H6O6} \)
(b) \( \ce{C6H14O12} \)
(c) \( \ce{C6H12O6} \)
(d) \( \ce{C6H14O6} \)
Show Answer & Explanation

Answer: (c) \( \ce{C6H12O6} \)

Explanation:
1. Empirical formula \( \ce{CH2O} \) (30 g/mol), so the molecular formula is (CH2O)n.
2. 178/30 = 5.9, so n = 6: \( \ce{C6H12O6} \) (180 g/mol).

Teacher's Note:
1. The molecular formula is a whole number multiple of the empirical formula.

Question 3: Air is taken as 80% \( \ce{N2} \) and 20% \( \ce{O2} \) by volume (so mole fractions are 0.8 and 0.2), and 1 mol of any gas occupies 22.4 L at STP. If air is treated as a solution of \( \ce{O2} \) and \( \ce{N2} \), the percentage by weight of oxygen is
(a) 10/9
(b) 200/9
(c) 700/9
(d) 350/9
Show Answer & Explanation

Answer: (b) 200/9

Explanation:
1. Per mole of air: \( \ce{O2} \) = 0.2 x 32 = 6.4 g; \( \ce{N2} \) = 0.8 x 28 = 22.4 g. Total = 28.8 g.
2. % W/W of oxygen = 6.4/28.8 x 100 = 22.2 = 200/9.

Teacher's Note:
1. Weight percent needs masses, not moles.

Question 4: Air is taken as 80% \( \ce{N2} \) and 20% \( \ce{O2} \) by volume (so mole fractions are 0.8 and 0.2), and 1 mol of any gas occupies 22.4 L at STP. The density of air at NTP is
(a) 1 g/L
(b) 2/7 g/L
(c) cannot be determined
(d) 9/7 g/L
Show Answer & Explanation

Answer: (d) 9/7 g/L

Explanation:
1. Mass of 1 mol of air = 28.8 g in 22.4 L.
2. Density = 28.8/22.4 = 9/7 g/L.

Teacher's Note:
1. Density = mass of a mole / molar volume.

Question 5: 60 g of a solution containing 40% by mass of NaCl is mixed with 100 g of a solution containing 15% by mass of NaCl. The molality of the final solution is (Na = 23, Cl = 35.5)
(a) 4.4 m
(b) 24.4 m
(c) 3.2 m
(d) 5.5 m
Show Answer & Explanation

Answer: (d) 5.5 m

Explanation:
1. NaCl = 39 g = 0.667 mol. Solvent = 160 - 39 = 121 g.
2. Molality = 0.667/0.121 = 5.5 m.

Teacher's Note:
1. Molality uses the mass of water only.

Question 6: 60 g of a solution containing 40% by mass of NaCl is mixed with 100 g of a solution containing 15% by mass of NaCl. The final solution has density 1.6 g/mL. Its molarity is (Na = 23, Cl = 35.5)
(a) 5.5 M
(b) 6.67 M
(c) 2.59 M
(d) 10.0 M
Show Answer & Explanation

Answer: (b) 6.67 M

Explanation:
1. NaCl = 39 g = 0.667 mol.
2. Volume = 160/1.6 = 100 mL = 0.1 L.
3. Molarity = 0.667/0.1 = 6.67 M.

Teacher's Note:
1. Use the density to find the volume of solution.

Question 7: A 4.0 M aqueous solution of NaCl (500 mL) is electrolysed, giving chlorine at the anode (\( \ce{2Cl- -> Cl2 + 2e-} \)) and sodium at the cathode (\( \ce{Na+ + e- -> Na} \); \( \ce{Na + Hg -> NaHg} \)). The total moles of chlorine gas evolved from complete electrolysis is
(a) 0.5
(b) 1.0
(c) 2.0
(d) 3.0
Show Answer & Explanation

Answer: (b) 1.0

Explanation:
1. NaCl = 4.0 x 0.5 = 2 mol, so Cl- = 2 mol.
2. \( \ce{2Cl- -> Cl2 + 2e-} \) gives 1 mol of \( \ce{Cl2} \).

Teacher's Note:
1. Two chloride ions give one chlorine molecule.

Question 8: A 4.0 M aqueous solution of NaCl (500 mL) is electrolysed with a mercury cathode. The maximum weight of sodium amalgam (NaHg) that can form is (Na = 23, Hg = 200)
(a) 200 g
(b) 446 g
(c) 225 g
(d) 400 g
Show Answer & Explanation

Answer: (b) 446 g

Explanation:
1. Na = 2 mol (46 g). NaHg has 1 : 1 stoichiometry, so Hg = 2 mol = 400 g.
2. Amalgam = 46 + 400 = 446 g.

Teacher's Note:
1. Add the mass of mercury that combines with the sodium.

Question 9: A 4.0 M aqueous solution of NaCl (500 mL) is electrolysed completely. The total charge required is (1 faraday = 96500 coulomb)
(a) 96500 C
(b) 193000 C
(c) 48250 C
(d) 24125 C
Show Answer & Explanation

Answer: (b) 193000 C

Explanation:
1. Cl- = 2 mol, each losing one electron, so 2 mol of electrons = 2 F.
2. Charge = 2 x 96500 = 193000 C.

Teacher's Note:
1. One mole of electrons carries one faraday.

Question 10: A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. The number of significant figures in the average titre value is
(a) 2
(b) 4
(c) 5
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Average = (25.2 + 25.25 + 25.0)/3 = 25.15 mL.
2. The least precise reading is 25.0 or 25.2 (3 significant figures), so the average is quoted to 3 significant figures.

Teacher's Note:
1. The result cannot be more precise than the least precise measurement.

Question 11: Reaction of \( Br_2 \) with \( Na_2CO_3 \) in aqueous solution gives sodium bromide and sodium bromate with evolution of \( CO_2 \). The number of sodium bromide molecules in the balanced chemical equation is
(a) 5
(b) 3
(c) 6
(d) 1
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. \( \ce{3Br2 + 3Na2CO3 -> 5NaBr + NaBrO3 + 3CO2} \).
2. Bromine: Br2 is reduced to Br- (5 e) and oxidised to BrO3- (5 e), so NaBr : NaBrO3 = 5 : 1.

Teacher's Note:
1. This is a disproportionation of bromine in basic medium.

Question 12: Dissolving 120 g of urea (molar mass 60) in 1000 g of water gives a solution of density 1.15 g/mL. The molarity of the solution is
(a) 1.78 M
(b) 2.00 M
(c) 2.22 M
(d) 2.05 M
Show Answer & Explanation

Answer: (d) 2.05 M

Explanation:
1. Urea = 2 mol. Mass of solution = 1120 g.
2. Volume = 1120/1.15 = 974 mL = 0.974 L.
3. Molarity = 2/0.974 = 2.05 M.

Teacher's Note:
1. Use the density to find the final volume of solution.

Question 13: A 29.2% (w/w) HCl stock solution has density 1.25 g/mL (molar mass 36.5). The volume of stock solution required to prepare 200 mL of 0.4 M HCl is
(a) 8 mL
(b) 16 mL
(c) 4 mL
(d) 80 mL
Show Answer & Explanation

Answer: (a) 8 mL

Explanation:
1. Molarity of the stock = 10 x 1.25 x 29.2/36.5 = 10 M.
2. V = (0.4 x 200)/10 = 8 mL.

Teacher's Note:
1. Use M1 V1 = M2 V2 after finding the stock molarity.

Question 14: The number of atoms in 560 g of Fe (atomic mass 56 g/mol) is
(a) twice that of 70 g of N
(b) half that of 20 g of H
(c) both (1) and (2)
(d) none of these
Show Answer & Explanation

Answer: (c) both (1) and (2)

Explanation:
1. Fe = 560/56 = 10 mol of atoms.
2. 70 g of N = 5 mol of atoms, so 10 mol is twice that. 20 g of H = 20 mol of atoms, so 10 mol is half that.
3. Both statements are correct.

Teacher's Note:
1. Compare in moles of atoms.

Question 15: In an organic compound of molar mass 108 g/mol, C, H and N atoms are present in the ratio 9 : 1 : 3.5 by weight. The molecular formula can be
(a) \( \ce{C6H8N2} \)
(b) \( \ce{C7H10N} \)
(c) \( \ce{C5H6N3} \)
(d) \( \ce{C4H18N3} \)
Show Answer & Explanation

Answer: (a) \( \ce{C6H8N2} \)

Explanation:
1. Moles: C = 9/12 = 0.75, H = 1, N = 3.5/14 = 0.25. Ratio 3 : 4 : 1.
2. Empirical formula \( \ce{C3H4N} \) = 54, so n = 2 and the formula is \( \ce{C6H8N2} \).

Teacher's Note:
1. Convert the weight ratio to a mole ratio and scale with the molar mass.

Question 16: Which of the following concentration terms is affected by a change in temperature?
(a) Molarity
(b) Molality
(c) Mole fraction
(d) Weight fraction
Show Answer & Explanation

Answer: (a) Molarity

Explanation:
1. Molarity involves the volume of solution, which changes with temperature.
2. The others use only masses or moles.

Teacher's Note:
1. Volume changes with temperature, mass does not.

Question 17: The volume of hydrogen gas at 273 K and 1 atm that will be consumed in obtaining 21.6 g of elemental boron (atomic mass 10.8) from the reduction of boron trichloride by hydrogen is
(a) 44.8 L
(b) 22.4 L
(c) 89.6 L
(d) 67.2 L
Show Answer & Explanation

Answer: (d) 67.2 L

Explanation:
1. \( \ce{2BCl3 + 3H2 -> 2B + 6HCl} \). B = 21.6/10.8 = 2 mol.
2. \( \ce{H2} \) = 3 mol = 67.2 L.

Teacher's Note:
1. Three moles of hydrogen are needed for every two moles of boron.

Question 18: If 1/6, in place of 1/12, of the mass of a carbon atom is taken as the relative atomic mass unit, the mass of one mole of a substance will
(a) decrease twice
(b) increase two-fold
(c) remain unchanged
(d) be a function of the molecular mass of the substance
Show Answer & Explanation

Answer: (c) remain unchanged

Explanation:
1. The mass of a mole is a physical quantity fixed by the definition of the mole in grams (the amount holding as many particles as 12 g of carbon-12).
2. Changing the unit of relative mass changes the numbers on the scale, but not the actual mass of a mole of the substance.

Teacher's Note:
1. A change of scale for relative masses does not change the physical mass of a mole.

Question 19: 480 mL of 1.5 M first solution of a substance is mixed with 520 mL of 1.2 M second solution (non-electrolyte). The molarity of the final mixture is
(a) 2.70 M
(b) 1.344 M
(c) 1.50 M
(d) 1.20 M
Show Answer & Explanation

Answer: (b) 1.344 M

Explanation:
1. Moles = 0.48 x 1.5 + 0.52 x 1.2 = 0.72 + 0.624 = 1.344 mol.
2. Volume = 1 L, so molarity = 1.344 M.

Teacher's Note:
1. Add the moles of solute and divide by the total volume.

Question 20: The number of moles of magnesium phosphate, \( Mg_3(PO_4)_2 \), that will contain 0.25 mol of oxygen atoms is
(a) 0.02
(b) 3.125 x \( 10^{-2} \)
(c) 1.25 x \( 10^{-2} \)
(d) 2.5 x \( 10^{-2} \)
Show Answer & Explanation

Answer: (b) 3.125 x \( 10^{-2} \)

Explanation:
1. Each formula unit has 8 O atoms.
2. Moles = 0.25/8 = 0.03125 = 3.125 x \( 10^{-2} \).

Teacher's Note:
1. Divide the moles of oxygen atoms by the number per formula unit.

Question 21: In the reaction \( \ce{2Al(s) + 6HCl(aq) -> 2Al^3+(aq) + 6Cl-(aq) + 3H2(g)} \),
(a) 6 L of HCl(aq) is consumed for every 3 L of \( H_2 \) produced
(b) 33.6 L of \( H_2 \) is produced regardless of temperature and pressure for every mole of Al that reacts
(c) 67.2 L of \( H_2 \) at STP is produced for every mole of Al that reacts
(d) 11.2 L of \( H_2 \) at STP is produced for every mole of HCl(aq) consumed
Show Answer & Explanation

Answer: (d) 11.2 L of \( H_2 \) at STP is produced for every mole of HCl(aq) consumed

Explanation:
1. 6 mol HCl gives 3 mol of \( \ce{H2} \), so 1 mol of HCl gives 0.5 mol = 11.2 L at STP.
2. One mole of Al gives 1.5 mol = 33.6 L at STP, and the volume depends on temperature and pressure. HCl(aq) is a solution, so a volume of gas cannot be compared with it.

Teacher's Note:
1. Use mole ratios and the molar volume at STP.

Question 22: The density (in g/mL) of a 3.60 M sulphuric acid solution that is 29% \( H_2SO_4 \) by mass (molar mass 98 g/mol) will be
(a) 1.22
(b) 1.45
(c) 1.64
(d) 1.88
Show Answer & Explanation

Answer: (a) 1.22

Explanation:
1. Density = molarity x molar mass/(10 x mass %) = 3.60 x 98/(10 x 29) = 1.22 g/mL.

Teacher's Note:
1. Use density = (M x molar mass)/(10 x percent).

Question 23: A 5.2 molal aqueous solution of methyl alcohol, \( \ce{CH3OH} \), is supplied. The mole fraction of methyl alcohol in the solution is
(a) 0.100
(b) 0.190
(c) 0.086
(d) 0.050
Show Answer & Explanation

Answer: (c) 0.086

Explanation:
1. 1 kg of water = 55.56 mol.
2. Mole fraction = 5.2/(5.2 + 55.56) = 0.086.

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

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

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