Mole Concept — Practice Questions

🧪 Mole Concept — Practice Questions

Tap a question to reveal the full step-by-step solution

🔵 Step 1: Molar Mass of NaOH

Na + O + H = 23 + 16 + 1 = 40 g mol⁻¹

🟢 Step 2: Moles of NaOH

Moles = Mass ÷ Molar Mass = 4 ÷ 40 = 0.1 mol

🟣 Step 3: Molarity of Original Solution

Molarity = Moles ÷ Volume (in L)

= 0.1 ÷ 0.250

= 0.4 mol L⁻¹ (0.4 M)

✅ Original Molarity = 0.4 M

🟠 Step 4: Molarity After Dilution to 1 L

Use the dilution formula (moles of solute stay the same):

M₁V₁ = M₂V₂

0.4 × 0.250 = M₂ × 1

M₂ = 0.1 mol L⁻¹

✅ New Molarity after dilution = 0.1 M

✅ Final Answer

Original Molarity = 0.4 M

Molarity after dilution to 1 L = 0.1 M

🔵 Step 1: Take Equal Mass "W" g for Both

Molar Mass of CO₂ = 12 + 32 = 44 g mol⁻¹

Molar Mass of O₂ = 2 × 16 = 32 g mol⁻¹

🟢 Step 2: Moles of Each Gas

Moles of CO₂ = W ÷ 44

Moles of O₂ = W ÷ 32

Since 44 > 32, dividing by the larger number gives fewer moles — so moles of CO₂ < moles of O₂ for the same mass.

🟣 Step 3: Compare Number of Molecules

Number of molecules = Moles × N_A

N(CO₂) : N(O₂) = (W/44) : (W/32) = 32 : 44 = 8 : 11

✅ O₂ sample has MORE molecules than CO₂ sample (ratio CO₂ : O₂ = 8 : 11)

🟠 Step 4: Compare Number of Oxygen Atoms

Each CO₂ molecule has 2 O atoms, and each O₂ molecule also has 2 O atoms — so the multiplying factor is the same for both:

O atoms in CO₂ = 2 × N(CO₂)

O atoms in O₂ = 2 × N(O₂)

Ratio = 2×(W/44) : 2×(W/32) = 8 : 11 (same as molecule ratio)

✅ O₂ sample also has MORE oxygen atoms (ratio CO₂ : O₂ = 8 : 11)

✅ Final Answer

Molecules ratio (CO₂ : O₂) = 8 : 11

Oxygen atoms ratio (CO₂ : O₂) = 8 : 11

(O₂ sample contains more molecules and more O atoms, since its molar mass is smaller)

🔵 Step 1: Oxygen Atoms per Molecule

1 molecule of H₂SO₄ contains 4 oxygen atoms

🟢 Step 2: Moles of Oxygen Atoms

Moles of O atoms = 0.25 × 4 = 1 mol

🟣 Step 3: Convert Moles to Number of Atoms

Number of atoms = Moles × N_A

= 1 × 6.022 × 10²³

= 6.022 × 10²³ atoms

✅ Final Answer

Number of oxygen atoms = 6.022 × 10²³

🔵 Step 1: Empirical Formula Mass

CH₂O = 12 + (2×1) + 16 = 30 g mol⁻¹

🟢 Step 2: Find "n" (Multiplying Factor)

n = Molar Mass ÷ Empirical Formula Mass

= 180 ÷ 30

n = 6

🟣 Step 3: Molecular Formula

Molecular Formula = (CH₂O)₆ = C₆H₁₂O₆

✅ Molecular formula = C₆H₁₂O₆ (this is glucose)

🟠 Step 4: Percentage of Oxygen

Mass of O in molecule = 6 × 16 = 96 g

% of O = (96 ÷ 180) × 100

= 53.33 %

✅ Final Answer

Molecular Formula = C₆H₁₂O₆

Percentage of Oxygen = 53.33 %

🔵 Step 1: Molar Mass of H₂O

(2×1) + 16 = 18 g mol⁻¹

🟢 Step 2: Moles of H₂O

Moles = 9 ÷ 18 = 0.5 mol

🟣 Step 3: Number of Molecules

Number of molecules = 0.5 × 6.022 × 10²³

= 3.011 × 10²³ molecules

✅ Number of H₂O molecules = 3.011 × 10²³

🟠 Step 4: Number of Hydrogen Atoms

Each H₂O molecule has 2 hydrogen atoms:

H atoms = 2 × 3.011 × 10²³

= 6.022 × 10²³ atoms

✅ Final Answer

Number of H₂O molecules = 3.011 × 10²³

Number of hydrogen atoms = 6.022 × 10²³