JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 08

Chemistry Objective Questions and Answers: Some Basic Concepts Of Chemistry

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Multiple Choice Questions

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

Answer: (c) both (a) and (b)

Explanation:
1. 560 g Fe = 10 mol
2. 70 g N = 5 mol of atoms, so Fe is twice
3. 20 g H = 20 mol, so Fe is half

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

Question 2: In an organic compound of molar mass 108 g/mol, C, H and N are present in the ratio 9 : 1 : 3.5 by mass. The molecular formula can be:
(a) C6H8N2
(b) C7H10N
(c) C5H6N3
(d) C4H18N3
Show Answer & Explanation

Answer: (a) C6H8N2

Explanation:
1. Moles: C 0.75, H 1, N 0.25 gives 3 : 4 : 1, empirical C3H4N (54)
2. Molar mass 108, so the formula is C6H8N2

Teacher's Note:
1. Divide by atomic masses and then by the smallest.

Question 3: When KMnO4 acts as an oxidising agent and ultimately forms MnO4(2-), MnO2, Mn2O3 and Mn2+, the number of electrons transferred in each case is respectively:
(a) 4, 3, 1, 5
(b) 1, 5, 3, 7
(c) 1, 3, 4, 5
(d) 3, 5, 7, 1
Show Answer & Explanation

Answer: (c) 1, 3, 4, 5

Explanation:
1. Mn(+7) to MnO4(2-) (+6): 1 e-
2. to MnO2 (+4): 3 e-
3. to Mn2O3 (+3): 4 e-
4. to Mn2+: 5 e-

Teacher's Note:
1. Count the drop in oxidation number from +7.

Question 4: Which of the following is a redox reaction?
(a) NaCl + KNO3 gives NaNO3 + KCl
(b) CaC2O4 + 2HCl gives CaCl2 + H2C2O4
(c) Mg(OH)2 + 2NH4Cl gives MgCl2 + 2NH4OH
(d) Zn + 2AgCN gives 2Ag + Zn(CN)2
Show Answer & Explanation

Answer: (d) Zn + 2AgCN gives 2Ag + Zn(CN)2

Explanation:
1. Only Zn + 2AgCN changes oxidation numbers (Zn 0 to +2, Ag +1 to 0)
2. The others are double decomposition

Teacher's Note:
1. Look for a change in oxidation number.

Question 5: 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 depends on volume, which changes with temperature
2. The others depend only on masses or moles

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

Question 6: What volume of hydrogen at 273 K and 1 atm is consumed in obtaining 21.6 g of elemental boron (atomic mass 10.8) by reducing BCl3 with hydrogen?
(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. 2BCl3 + 3H2 gives 2B + 6HCl
2. 21.6 g B = 2 mol, needing 3 mol H2
3. Volume = 3 x 22.4 = 67.2 L

Teacher's Note:
1. Use the balanced equation ratio.

Question 7: 6.02 x 10^20 molecules of urea are present in 100 mL of its solution. The concentration of the solution is:
(a) 0.001 M
(b) 0.01 M
(c) 0.02 M
(d) 0.1 M
Show Answer & Explanation

Answer: (b) 0.01 M

Explanation:
1. Moles = 6.02e20/6.02e23 = 1e-3
2. Molarity = 1e-3/0.1 L = 0.01 M

Teacher's Note:
1. Divide moles by litres.

Question 8: The oxidation state of Cr in [Cr(NH3)4Cl2]+ is:
(a) +3
(b) +2
(c) +1
(d) 0
Show Answer & Explanation

Answer: (a) +3

Explanation:
1. NH3 is neutral and each Cl- is -1
2. x + 0 - 2 = +1, so x = +3

Teacher's Note:
1. Charge of complex = sum of oxidation numbers.

Question 9: The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate is:
(a) +4
(b) +6
(c) +2
(d) +3
Show Answer & Explanation

Answer: (d) +3

Explanation:
1. Cr2O7(2-) oxidises I- to I2 and is reduced to Cr3+

Teacher's Note:
1. Dichromate in acid gives Cr(III).

Question 10: 480 mL of 1.5 M of a non-electrolyte solution is mixed with 520 mL of 1.2 M of the same solute. 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 = 0.72 + 0.624 = 1.344
2. Volume = 1.0 L
3. Molarity = 1.344 M

Teacher's Note:
1. Add moles and volumes.

Question 11: Which reaction depicts the oxidising behaviour of H2SO4?
(a) 2HI + H2SO4 gives I2 + SO2 + 2H2O
(b) Ca(OH)2 + H2SO4 gives CaSO4 + 2H2O
(c) NaCl + H2SO4 gives NaHSO4 + HCl
(d) 2PCl5 + H2SO4 gives 2POCl3 + 2HCl + SO2Cl2
Show Answer & Explanation

Answer: (a) 2HI + H2SO4 gives I2 + SO2 + 2H2O

Explanation:
1. In the first reaction S(+6) is reduced to S(+4) in SO2
2. The other reactions leave S at +6

Teacher's Note:
1. Look for the sulphur being reduced.

Question 12: How many moles of magnesium phosphate, Mg3(PO4)2, will contain 0.25 mole of oxygen atoms?
(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. One mole of Mg3(PO4)2 holds 8 mol O
2. Moles = 0.25/8 = 3.125 x 10^-2

Teacher's Note:
1. Divide by oxygen atoms per formula unit.

Question 13: The density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is:
(a) 2.28 mol/kg
(b) 3.28 mol/kg
(c) 1.14 mol/kg
(d) 0.44 mol/kg
Show Answer & Explanation

Answer: (a) 2.28 mol/kg

Explanation:
1. Per litre: mass 1020 g, acetic acid 2.05 x 60 = 123 g
2. Water = 897 g
3. Molality = 2.05/0.897 = 2.28

Teacher's Note:
1. Molality needs the mass of solvent.

Question 14: In the reaction 2Al(s) + 6HCl(aq) gives 2Al3+ + 6Cl- + 3H2(g):
(a) 6 L HCl(aq) is consumed for every 3 L H2 produced
(b) 33.6 L H2 is produced regardless of temperature and pressure for every mole of Al that reacts
(c) 67.2 L H2 at STP is produced for every mole of Al that reacts
(d) 11.2 L H2 at STP is produced for every mole of HCl consumed
Show Answer & Explanation

Answer: (d) 11.2 L H2 at STP is produced for every mole of HCl consumed

Explanation:
1. One mole Al gives 1.5 mol H2 = 33.6 L at STP only, so (b) and (c) are wrong
2. 6 mol HCl gives 3 mol H2, so 1 mol HCl gives 0.5 mol = 11.2 L at STP

Teacher's Note:
1. Check the mole ratios.

Question 15: The density (g/mL) of a 3.60 M sulphuric acid solution that is 29% H2SO4 by mass (molar mass 98) is:
(a) 1.22
(b) 1.45
(c) 1.64
(d) 1.88
Show Answer & Explanation

Answer: (a) 1.22

Explanation:
1. 1 L contains 3.6 x 98 = 352.8 g H2SO4, which is 29%
2. Mass of 1 L = 352.8/0.29 = 1216.6 g
3. Density = 1.22 g/mL

Teacher's Note:
1. Use mass of solute and mass percent.

Question 16: A 5.2 molal aqueous solution of methyl alcohol is supplied. The mole fraction of methyl alcohol is:
(a) 0.100
(b) 0.190
(c) 0.086
(d) 0.050
Show Answer & Explanation

Answer: (c) 0.086

Explanation:
1. 1 kg water = 55.56 mol
2. x = 5.2/(5.2 + 55.56) = 0.0856

Teacher's Note:
1. Molal means moles per kg of water.

Question 17: The molality of a solution of 0.0100 g of urea in 0.3000 dm3 of water is:
(a) 5.55 x 10^-4 m
(b) 33.3 m
(c) 3.33 x 10^-2 m
(d) 0.555 m
Show Answer & Explanation

Answer: (a) 5.55 x 10^-4 m

Explanation:
1. Moles urea = 0.01/60 = 1.667 x 10^-4
2. Water = 0.300 kg
3. Molality = 5.55 x 10^-4

Teacher's Note:
1. 1 dm3 of water is about 1 kg.

Question 18: 120 g of urea (molar mass 60) is dissolved in 1000 g of water, giving a solution of density 1.15 g/mL. The molarity is:
(a) 0.50 M
(b) 1.78 M
(c) 1.02 M
(d) 2.05 M
Show Answer & Explanation

Answer: (d) 2.05 M

Explanation:
1. Moles = 2
2. Volume = 1120/1.15 = 973.9 mL
3. Molarity = 2/0.9739 = 2.05 M

Teacher's Note:
1. Volume comes from mass divided by density.

Question 19: 1.4 g of an organic compound is digested by Kjeldahl's method and the ammonia is absorbed in 60 mL of M/10 H2SO4. The unreacted acid needs 20 mL of M/10 NaOH. The percentage of nitrogen is:
(a) 3%
(b) 5%
(c) 6%
(d) 10%
Show Answer & Explanation

Answer: (d) 10%

Explanation:
1. Acid taken = 12 meq; NaOH = 2 meq, so NH3 absorbed = 10 meq
2. N = 10 x 14 = 140 mg
3. 140/1400 = 10%

Teacher's Note:
1. H2SO4 is dibasic, so 60 mL of M/10 is 12 meq.

Question 20: The molecular formula of a resin used for water softening is C8H7SO3Na (molar mass 206). The maximum uptake of Ca2+ per gram of resin, in mole, is:
(a) 2/309
(b) 1/412
(c) 1/103
(d) 1/206
Show Answer & Explanation

Answer: (b) 1/412

Explanation:
1. One gram holds 1/206 mol Na+
2. Each Ca2+ replaces 2 Na+
3. Uptake = 1/412 mol

Teacher's Note:
1. Charge balance: one Ca2+ for two Na+.

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