Question 1

Match List-I with List-II. List-I Mass of substance List-II Number of atoms A. 1.8 mg water I. 2 10^ -4 N_A B. 9.8 mg sulphuric acid II. 1.5 10^ -4 N_A C. 1.8 mg carbon III. 3 10^ -4 N_A D. 5.85 mg sa

Accepted Answer

For A: Molar mass of H_2O = 18 g/mol. Moles of H_2O = 1.8 10^ -3 18 = 10^ -4 . Since each H_2O molecule has 3 atoms, the number of atoms = 3 10^ -4 N_A. For B: Molar mass of H_2SO_4 = 98 g/mol. Moles of H_2SO_4 = 9.8 10^ -3 98 = 10^ -4 . Since each H_2SO_4 molecule has 7 atoms, the number of atoms = 7 10^ -4 N_A. For C: Molar mass of C = 12 g/mol. Moles of C = 1.8 10^ -3 12 = 1.5 10^ -4 . The number of atoms = 1.5 10^ -4 N_A. For D: Molar mass of NaCl = 58.5 g/mol. Moles of NaCl = 5.85 10^ -3 58

Question 2

Which of the following contain the same number of atoms ? (Given : Molar mass in g mol^ -1 of H, He, O and S are 1, 4, 16 and 32 respectively) A. 2 g of O_2 gas B. 4 g of SO_2 gas C. 1400 mL of O_2 at

Accepted Answer

Number of atoms is calculated as the product of the number of moles, atomicity, and Avogadro's number (N_A). For A: Moles of O_2 = 2 32 = 1 16 mol. Atomicity of O_2 is 2. Number of atoms = 1 16 2 N_A = 1 8 N_A For B: Moles of SO_2 = 4 64 = 1 16 mol. Atomicity of SO_2 is 3. Number of atoms = 1 16 3 N_A = 3 16 N_A For C: Moles of O_2 at STP = 1400 22400 = 1 16 mol. Atomicity of O_2 is 2. Number of atoms = 1 16 2 N_A = 1 8 N_A For D: Moles of He at STP = 0.05 22.4 = 1 448 mol. Atomicity of He is 1.

Question 3

An oxide of iron contains 69.9\% iron, its empirical formula, is: (Given: Molar mass of Fe and O are 56 and 16 g mol^ -1 respectively.)

Accepted Answer

Mass percentage of Fe = 69.9\% Mass percentage of O = 100\% - 69.9\% = 30.1\% Number of moles of Fe = 69.9 56 = 1.248 Number of moles of O = 30.1 16 = 1.881 Molar ratio of Fe : O = 1.248 : 1.881 Dividing by the smallest value, the ratio is 1.248 1.248 : 1.881 1.248 = 1 : 1.5 Multiplying by 2 to get the simplest whole number ratio, we get 2 : 3. The empirical formula is Fe _2 O _3. Answer: Fe _2 O _3

Question 4

What volume of hydrogen gas at STP would be liberated by action of 50 mL of H_2SO_4 of 50\% purity (density = 1.3 g mL ^ -1 ) on 20 g of zinc ? Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mo

Accepted Answer

Mass of H_2SO_4 solution = 50 1.3 = 65 g Mass of pure H_2SO_4 = 65 50 100 = 32.5 g Moles of H_2SO_4 = 32.5 98 0.3316 mol Moles of Zn = 20 65 0.3077 mol The balanced chemical equation is: Zn + H_2SO_4 ZnSO_4 + H_2 Since the moles of Zn are less than the moles of H_2SO_4, Zn is the limiting reagent. Moles of H_2 produced = Moles of Zn = 20 65 mol Volume of H_2 gas at STP = 20 65 22.4 L = 6.892 L Answer: 6.892 L

Question 5

How many grams of residue is obtained by heating 2.76 g of silver carbonate? (Given: Molar mass of C, O and Ag are 12, 16 and 108 g mol^ -1 respectively)

Accepted Answer

Molar mass of Ag_2CO_3 = 2 108 + 12 + 3 16 = 276 g mol^ -1 Number of moles of Ag_2CO_3 = 2.76 276 = 0.01 mol The decomposition reaction of silver carbonate on heating is: Ag_2CO_3(s) 2Ag(s) + CO_2(g) + 1 2 O_2(g) From the stoichiometry of the reaction, 1 mole of Ag_2CO_3 gives 2 moles of Ag as residue. Number of moles of Ag formed = 2 0.01 = 0.02 mol Mass of Ag residue = 0.02 108 = 2.16 g Answer: 2.16 g

Question 6

The correct order of total number of atoms present in (A) 2 moles of cyclohexane (B) 684 g of sucrose (C) 90.8 L of dihydrogen at STP is:

Accepted Answer

For (A): Formula of cyclohexane is C_6H_ 12 . Number of atoms per molecule = 6 + 12 = 18. Total number of atoms in 2 moles = 2 18 N_A = 36 N_A. For (B): Formula of sucrose is C_ 12 H_ 22 O_ 11 . Molar mass of sucrose = 342 g/mol. Number of moles = 684 342 = 2 moles. Number of atoms per molecule = 12 + 22 + 11 = 45. Total number of atoms = 2 45 N_A = 90 N_A. For (C): Formula of dihydrogen is H_2. Number of moles at STP = 90.8 22.7 = 4 moles. Number of atoms per molecule = 2. Total number of atoms

Question 7

Number of moles and number of molecules in 1.4187 L of SO_2 at STP respectively are:

Accepted Answer

Volume of SO_2 at STP = 1.4187 L Molar volume of an ideal gas at STP is approximately 22.414 L. Number of moles n = Volume Molar Volume = 1.4187 22.414 = 0.0633 Number of molecules = n N_A Number of molecules = 0.0633 6.022 10^ 23 = 3.812 10^ 22 Answer: 0.0633; 3.812 10^ 22

Question 8

The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1. It has two carbon atoms. The weight (in g) of CO_2(g) formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is :

Accepted Answer

Ratio of mass percentage of C : H = 12 : 1 Ratio of moles of C : H = 12 12 : 1 1 = 1 : 1 The empirical formula is CH. Since the hydrocarbon has two carbon atoms, its molecular formula is C_2H_2. Molar mass of C_2H_2 = 2 12 + 2 1 = 26 g mol^ -1 Moles of C_2H_2 in 3.38 g = 3.38 26 = 0.13 mol The combustion reaction is: C_2H_2 + 5 2 O_2 2CO_2 + H_2O From the stoichiometry, 1 mole of C_2H_2 produces 2 moles of CO_2. Moles of CO_2 produced = 2 0.13 = 0.26 mol Molar mass of CO_2 = 12 + 2 16 = 44 g mol

Question 9

For the given reaction: CaCO _ 3 +2 HCl CaCl _ 2 + H _ 2 O + CO _ 2 If 90 ~g CaCO _ 3 is added to 300 mL of HCl which contains 38.55 \% HCl by mass and has density 1.13 ~g ~mL ^ -1 , then which of the

Accepted Answer

Molar masses: HCl = 36.5, CaCO₃ = 100 g/mol HCl solution: mass = 300 × 1.13 = 339 g HCl present: 339 × 0.3855 = 130.74 g = 130.74/36.5 = 3.58 mol CaCO₃: 90/100 = 0.9 mol Reaction: CaCO_3 + 2HCl → CaCl_2 + H_2O + CO_2 HCl required: 0.9 × 2 = 1.8 mol HCl available: 3.58 mol > 1.8 mol CaCO₃ is limiting. HCl remaining: (3.58 - 1.8) × 36.5 = 1.78 × 36.5 = 64.97 g

Question 10

A +2 ~B AB _ 2 36.0 g of 'A' (Molar mass: 60 ~g ~mol ^ -1 ) and 56.0 g of ' B ^ (Molar mass: 80 ~g ~mol ^ -1 ) are allowed to react. Which of the following statements are correct ? A. 'A' is the limit

Accepted Answer

Reaction: A + 2B → AB₂. Moles of A = 36.0/60 = 0.6 mol; Moles of B = 56.0/80 = 0.7 mol. A requires 1.2 mol of B but only 0.7 available, so B is limiting reagent. So, (A) False - B is limiting. (B) AB₂ formed = 0.7/2 = 0.35 mol; Molar mass AB₂ = 60 + 2(80) = 220 g/mol; Mass = 0.35 × 220 = 77 g True. (C) False - Molar mass is 220 g/mol. (D) A remaining = 0.6 - 0.35 = 0.25 mol = 15 g True. Correct statements: B and D.

Question 11

In the reaction, 2 Al ( s )+6 HCl ( aq ) 2 Al ^ 3+ ( aq )+6 Cl ^ - ( aq )+3 H _ 2 ( ~g )

Accepted Answer

Reaction: 2 Al (s) + 6 HCl (aq) 2 Al ^ 3+ (aq) + 6 Cl ^-(aq) + 3 H _2(g) Checking each option: (1) Per 1 mole HCl: From stoichiometry, 6 mol HCl produces 3 mol H₂, so 1 mol HCl produces 0.5 mol H₂ = 11.2 L at STP. TRUE (2) Per 1 mole Al: 2 mol Al produces 3 mol H₂, so 1 mol Al produces 1.5 mol H₂ = 33.6 L at STP only, not regardless of T and P. FALSE (3) Volume ratio at same T,P: 6 mol HCl : 3 mol H₂ = 2:1, so 12 L HCl : 6 L H₂. TRUE (4) Per 1 mole Al: 1.5 mol H₂ = 33.6 L, not 67.2 L. FALSE Opti

Question 12

Aqueous HCl reacts with MnO _ 2 ( ~s ) to form MnCl _ 2 ( aq ), Cl _ 2 ( ~g ) and H _ 2 O (l). What is the weight (in g) of Cl _ 2 liberated when 8.7 g of MnO _ 2 ( ~s ) is reacted with excess aqueous

Accepted Answer

The balanced redox equation is: MnO _2 + 4 HCl MnCl _2 + Cl _2 + 2 H _2 O . Molar mass of MnO₂ = 55 + 32 = 87 g/mol. Moles of MnO₂ = 8.7 87 = 0.1 mol. From stoichiometry, 1 mol MnO₂ produces 1 mol Cl₂, so 0.1 mol produces 0.1 mol Cl₂. Mass of Cl₂ = 0.1 71 = 7.1 g.

Question 13

14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect? [Given : Molar mass in g mol ^ -1 of Ca -40, Cl -35.5, H -1]

Accepted Answer

Reaction: Ca + 2HCl CaCl_2 + H_2 Moles of Ca = 14.0 40 = 0.35 mol. Since HCl is in excess, Ca is limiting. Option (1): H_2 evolved = 0.35 mol Option (2): Volume of H_2 at STP = 0.35 22.4 = 7.84 L Option (3): Ca is limiting reagent Option (4): Mass of CaCl_2 = 0.35 111 = 38.85 g, not 33.3 g Statement (4) is incorrect.

Question 14

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH so

Accepted Answer

KOH absorbs CO_2. Volume of CO_2 produced = 224 - 64 = 160 mL Unreacted O_2 = 64 mL, so O_2 consumed = 264 - 64 = 200 mL For hydrocarbon C_xH_y: CO_2 Hydrocarbon = 160 80 = 2 x = 2 O_2 consumed per mole = x + y 4 = 200 80 = 2.5 2 + y 4 = 2.5 y = 2 Hydrocarbon is C_2H_2

Some Basic Concepts of Chemistry — Previous Year Questions & Solutions