Some Basic Concepts of Chemistry

🧪 Some Basic Concepts of Chemistry

JEE Level — 6 Marks Questions — Tap a question to reveal full solution

🔵 Step 1: Moles of Each Gas

Moles CO₂ = 4.4 ÷ 44 = 0.1 mol

Moles O₂ = 3.2 ÷ 32 = 0.1 mol

Moles H₂ = 2.0 ÷ 2 = 1.0 mol

🟢 Step 2: Total Moles

Total moles = 0.1 + 0.1 + 1.0 = 1.2 mol

🟣 Step 3: Total Number of Molecules

1.2 × 6.022 × 10²³ = 7.226 × 10²³ molecules

🟠 Step 4: Mole Fraction of Each Gas

x(CO₂) = 0.1 ÷ 1.2 = 0.083

x(O₂) = 0.1 ÷ 1.2 = 0.083

x(H₂) = 1.0 ÷ 1.2 = 0.833

✅ Final Answer

Total molecules = 7.226 × 10²³

Mole fractions: CO₂ = 0.083, O₂ = 0.083, H₂ = 0.833

🔵 Step 1: Mass of Carbon (from CO₂)

Moles CO₂ = 0.88 ÷ 44 = 0.02 mol = moles C

Mass C = 0.02 × 12 = 0.24 g

🟢 Step 2: Mass of Hydrogen (from H₂O)

Moles H₂O = 0.36 ÷ 18 = 0.02 mol → moles H = 0.04 mol

Mass H = 0.04 × 1 = 0.04 g

🟣 Step 3: Mass of Oxygen (by difference)

Mass O = 0.44 − 0.24 − 0.04 = 0.16 g

Moles O = 0.16 ÷ 16 = 0.01 mol

🟠 Step 4: Mole Ratio

C : H : O = 0.02 : 0.04 : 0.01 = 2 : 4 : 1

✅ Final Answer

Empirical Formula = C₂H₄O

🔵 Step 1: Moles of Each Reactant

Moles Al = 5.4 ÷ 27 = 0.2 mol

Moles Cl₂ = 14.2 ÷ 71 = 0.2 mol

🟢 Step 2: Compare With Stoichiometric Ratio (2 : 3)

0.2 mol Al needs (3/2)×0.2 = 0.3 mol Cl₂

Only 0.2 mol Cl₂ available → Cl₂ is limiting

✅ Limiting Reagent = Cl₂

🟣 Step 3: Moles of AlCl₃ Formed

Using Cl₂: moles AlCl₃ = (2/3) × 0.2 = 0.1333 mol

🟠 Step 4: Mass of AlCl₃

Molar mass AlCl₃ = 27 + (3×35.5) = 133.5 g/mol

Mass = 0.1333 × 133.5 = 17.8 g

✅ Final Answer

Limiting Reagent = Cl₂

Mass of AlCl₃ formed = 17.8 g

🔵 Step 1: Reaction on Heating

Only NaHCO₃ decomposes on heating:

2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

🟢 Step 2: Moles of CO₂ and NaHCO₃

Moles CO₂ = 0.56 ÷ 44 = 0.01273 mol

Moles NaHCO₃ = 2 × 0.01273 = 0.02545 mol

🟣 Step 3: Mass of NaHCO₃ and Na₂CO₃

Mass NaHCO₃ = 0.02545 × 84 = 2.138 g

Mass Na₂CO₃ = 2.50 − 2.138 = 0.362 g

🟠 Step 4: Percentage Composition

% NaHCO₃ = (2.138/2.50) × 100 = 85.5%

% Na₂CO₃ = 14.5%

✅ Final Answer

NaHCO₃ = 85.5%  |  Na₂CO₃ = 14.5%

🔵 Step 1: Set Up Equations

Let x = mass CaCO₃, y = mass MgCO₃; x + y = 10

CaCO₃ → CaO + CO₂ (loses 44/100 of mass)
MgCO₃ → MgO + CO₂ (loses 44/84 of mass)

🟢 Step 2: Mass Loss Equation

Total mass loss = 10.0 − 5.20 = 4.80 g

0.44x + 0.5238(10−x) = 4.80

🟣 Step 3: Solve for x and y

−0.0838x = −0.438 → x = 5.23 g CaCO₃

y = 10 − 5.23 = 4.77 g MgCO₃

✅ Final Answer

CaCO₃ = 52.3%  |  MgCO₃ = 47.7%

🔵 Step 1: Moles of KClO₃

Molar mass KClO₃ = 39+35.5+48 = 122.5 g/mol

Moles = 5.00 ÷ 122.5 = 0.0408 mol

🟢 Step 2: Moles of O₂ (ratio 2:3)

Moles O₂ = (3/2) × 0.0408 = 0.0612 mol

🟣 Step 3: Volume of O₂ at STP

Volume = 0.0612 × 22.4 = 1.371 L

🟠 Step 4: Number of O₂ Molecules

0.0612 × 6.022×10²³ = 3.687 × 10²² molecules

✅ Final Answer

Volume of O₂ = 1.371 L  |  Molecules = 3.687 × 10²²

🔵 Step 1: Mole Ratio from % Composition

C = 85.7 ÷ 12 = 7.14

H = 14.3 ÷ 1 = 14.3

Ratio C : H = 7.14 : 14.3 = 1 : 2

✅ Empirical Formula = CH₂

🟢 Step 2: Molar Mass from Vapour Density

Molar Mass = 2 × Vapour Density = 2 × 21 = 42 g/mol

🟣 Step 3: Find n

Empirical formula mass (CH₂) = 14

n = 42 ÷ 14 = 3

✅ Final Answer

Empirical Formula = CH₂

Molecular Formula = C₃H₆

🔵 Step 1: Total Moles of CO₂

Moles CO₂ = 1.12 ÷ 22.4 = 0.05 mol

🟢 Step 2: Set Up Equations

Let x = mol Na₂CO₃, y = mol NaHCO₃ (both give 1:1 CO₂)

x + y = 0.05

106x + 84y = 5.00

🟣 Step 3: Solve

106(0.05−y) + 84y = 5.00 → −22y = −0.30

y = 0.0136 mol (NaHCO₃), x = 0.0364 mol (Na₂CO₃)

🟠 Step 4: Convert to Mass

Mass Na₂CO₃ = 0.0364 × 106 = 3.86 g

Mass NaHCO₃ = 0.0136 × 84 = 1.14 g

✅ Final Answer

Na₂CO₃ = 3.86 g (77.1%)  |  NaHCO₃ = 1.14 g (22.9%)

🔵 Step 1: Moles of Each Reactant

Moles CaCO₃ = 10 ÷ 100 = 0.1 mol

Moles HCl = 0.100 L × 1.0 M = 0.1 mol

🟢 Step 2: Compare With Ratio (1:2)

0.1 mol CaCO₃ needs 0.2 mol HCl, only 0.1 mol available

✅ Limiting Reagent = HCl

🟣 Step 3: Moles of CO₂ Formed

Moles CO₂ = moles HCl ÷ 2 = 0.1 ÷ 2 = 0.05 mol

🟠 Step 4: Volume at STP

Volume = 0.05 × 22.4 = 1.12 L

✅ Final Answer

Limiting Reagent = HCl

Volume of CO₂ = 1.12 L

🔵 Step 1: Moles of NaCl

Molar mass NaCl = 58.5 g/mol

Moles = 5.85 ÷ 58.5 = 0.1 mol

🟢 Step 2: Molality

Molality = moles solute ÷ kg solvent = 0.1 ÷ 0.500 = 0.2 m

🟣 Step 3: Moles of Water

Moles H₂O = 500 ÷ 18 = 27.78 mol

🟠 Step 4: Mole Fraction of NaCl

x(NaCl) = 0.1 ÷ (0.1+27.78) = 3.59 × 10⁻³

✅ Final Answer

Molality = 0.2 m  |  Mole fraction of NaCl = 3.59 × 10⁻³

🔵 Step 1: Moles of CO₂

Moles CO₂ = 3.36 ÷ 22.4 = 0.15 mol

🟢 Step 2: Moles of CaCO₃ (1:1 ratio)

Moles CaCO₃ = 0.15 mol

🟣 Step 3: Mass of CaCO₃

Mass = 0.15 × 100 = 15 g

🟠 Step 4: Percentage

% CaCO₃ = (15 ÷ 20) × 100 = 75%

✅ Final Answer

Percentage of CaCO₃ = 75%

🔵 Step 1: Mole Ratio

C = 92.3 ÷ 12 = 7.69

H = 7.7 ÷ 1 = 7.7

Ratio C:H = 1:1

✅ Empirical Formula = CH

🟢 Step 2: Empirical Formula Mass

CH = 12 + 1 = 13 g/mol

🟣 Step 3: Find n

n = 78 ÷ 13 = 6

✅ Final Answer

Empirical Formula = CH

Molecular Formula = C₆H₆ (Benzene)

🔵 Step 1: Moles of Each Reactant

Moles H₂ = 2.0 ÷ 2 = 1.0 mol

Moles O₂ = 16 ÷ 32 = 0.5 mol

🟢 Step 2: Compare With Ratio (2:1)

1.0 mol H₂ needs exactly 0.5 mol O₂ — this exactly matches what is available!

✅ Reactants are in exact stoichiometric ratio — no limiting reagent, both fully consumed

🟣 Step 3: Mass of Water Formed

Moles H₂O = moles H₂ = 1.0 mol

Mass = 1.0 × 18 = 18 g

✅ Final Answer

No limiting reagent (exact stoichiometric mixture)

Mass of H₂O formed = 18 g

Excess reactant remaining = 0 g

🔵 Step 1: Reaction and Mass Loss

2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Mass loss = 10 − 6.90 = 3.10 g (as CO₂+H₂O)

🟢 Step 2: Moles of NaHCO₃

2 mol NaHCO₃ (168 g) loses 62 g (CO₂+H₂O)

Let n = mol NaHCO₃; loss = 31n = 3.10 → n = 0.1 mol

🟣 Step 3: Mass of NaHCO₃ and Na₂CO₃

Mass NaHCO₃ = 0.1 × 84 = 8.4 g

Mass Na₂CO₃ (original) = 10 − 8.4 = 1.6 g

🟠 Step 4: Verify Residue

Residue = 1.6 + (0.05 × 106) = 1.6 + 5.3 = 6.9 g ✓ matches given data

✅ Final Answer

NaHCO₃ = 84%  |  Na₂CO₃ = 16%

🔵 Step 1: Mass of Carbon and Hydrogen

Moles CO₂ = 4.4÷44 = 0.1 mol → mass C = 0.1×12 = 1.2 g

Moles H₂O = 1.8÷18 = 0.1 mol → moles H = 0.2 mol → mass H = 0.2 g

🟢 Step 2: Mass of Oxygen (by difference)

Mass O = 2.0 − 1.2 − 0.2 = 0.6 g → moles O = 0.6÷16 = 0.0375 mol

🟣 Step 3: Mole Ratio

C : H : O = 0.1 : 0.2 : 0.0375 = 8 : 16 : 3 (dividing by 0.0375, ×3)

⚠️ Note: this gives empirical formula mass ≈ 160 g/mol, which is already greater than the given molar mass of 60 g/mol. This means the numbers in the question are not mutually consistent (a real compound's molecular mass can never be smaller than its empirical formula mass). The method above is the correct approach — with consistent data you would proceed to find n = Molar Mass ÷ Empirical Formula Mass and multiply the empirical formula by n to get the molecular formula.

✅ Working Shown

% composition: C = 60%, H = 10%, O = 30% by mass

Empirical formula (as calculated) = C₈H₁₆O₃

Please double-check the given data (mass of sample / CO₂ / H₂O / molar mass) — as stated, it does not correspond to a valid molecular formula at M = 60 g/mol.