SOLUTION: Mole concept questions - Studypool - Free Printable

Section (A): Laws of chemical combination, atoms, molecules, moles & Avogadro's hypothesis

#### A-1. If the atomic mass of Sodium is 23, the number of moles in 46 g of sodium is:
- Given: Atomic mass of sodium = 23 g/mol
- Mass of sodium = 46 g
- Formula for moles:
\[
\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}}
\]
- Calculation:
\[
\text{Number of moles} = \frac{46 \, \text{g}}{23 \, \text{g/mol}} = 2 \, \text{moles}
\]

- Answer: (2) 2

---

#### A-2. Which of the following contains the greatest number of atoms?
- Given options:
1. 1.0 g of butane (\(C_4H_{10}\))
2. 1.0 g of nitrogen (\(N_2\))
3. 1.0 g of silver (Ag)
4. 1.0 g of water (\(H_2O\))

- Step 1: Calculate the number of moles for each substance.
- Butane (\(C_4H_{10}\)):
- Molar mass of \(C_4H_{10}\) = \(4 \times 12 + 10 \times 1 = 58 \, \text{g/mol}\)
- Moles of butane = \(\frac{1.0 \, \text{g}}{58 \, \text{g/mol}} \approx 0.0172 \, \text{mol}\)
- Number of atoms per molecule = \(4 + 10 = 14\)
- Total atoms = \(0.0172 \, \text{mol} \times 14 \times 6.022 \times 10^{23} \approx 1.45 \times 10^{23}\)

- Nitrogen (\(N_2\)):
- Molar mass of \(N_2\) = \(2 \times 14 = 28 \, \text{g/mol}\)
- Moles of \(N_2\) = \(\frac{1.0 \, \text{g}}{28 \, \text{g/mol}} \approx 0.0357 \, \text{mol}\)
- Number of atoms per molecule = 2
- Total atoms = \(0.0357 \, \text{mol} \times 2 \times 6.022 \times 10^{23} \approx 4.30 \times 10^{22}\)

- Silver (Ag):
- Molar mass of Ag = \(108 \, \text{g/mol}\)
- Moles of Ag = \(\frac{1.0 \, \text{g}}{108 \, \text{g/mol}} \approx 0.00926 \, \text{mol}\)
- Number of atoms per mole = \(6.022 \times 10^{23}\)
- Total atoms = \(0.00926 \, \text{mol} \times 6.022 \times 10^{23} \approx 5.57 \times 10^{21}\)

- Water (\(H_2O\)):
- Molar mass of \(H_2O\) = \(2 \times 1 + 16 = 18 \, \text{g/mol}\)
- Moles of \(H_2O\) = \(\frac{1.0 \, \text{g}}{18 \, \text{g/mol}} \approx 0.0556 \, \text{mol}\)
- Number of atoms per molecule = 3
- Total atoms = \(0.0556 \, \text{mol} \times 3 \times 6.022 \times 10^{23} \approx 1.01 \times 10^{23}\)

- Comparison:
- Butane: \(1.45 \times 10^{23}\) atoms
- Nitrogen: \(4.30 \times 10^{22}\) atoms
- Silver: \(5.57 \times 10^{21}\) atoms
- Water: \(1.01 \times 10^{23}\) atoms

- Answer: (1) 1.0 g of butane (\(C_4H_{10}\))

---

#### A-3. A sample of aluminium has a mass of 54.0 g. What is the mass of the same number of magnesium atoms?
- Given: Mass of aluminium = 54.0 g, Atomic mass of Al = 27 g/mol, Atomic mass of Mg = 24 g/mol
- Step 1: Calculate the number of moles of aluminium.
\[
\text{Moles of Al} = \frac{54.0 \, \text{g}}{27 \, \text{g/mol}} = 2 \, \text{mol}
\]
- Step 2: Since the number of atoms is the same, the number of moles of magnesium must also be 2 mol.
- Step 3: Calculate the mass of 2 moles of magnesium.
\[
\text{Mass of Mg} = 2 \, \text{mol} \times 24 \, \text{g/mol} = 48 \, \text{g}
\]

- Answer: (3) 48 g

---

#### A-4. The weight of a molecule of the compound \(C_6H_{12}O_6\) is:
- Given: Compound = \(C_6H_{12}O_6\)
- Step 1: Calculate the molar mass of \(C_6H_{12}O_6\).
\[
\text{Molar mass} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \, \text{g/mol}
\]
- Step 2: Calculate the mass of one molecule using Avogadro's number.
\[
\text{Mass of one molecule} = \frac{\text{Molar mass}}{\text{Avogadro's number}} = \frac{180 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 2.988 \times 10^{-22} \, \text{g}
\]

- Answer: (2) \(2.988 \times 10^{-22} \, \text{g}\)

---

#### A-5. Four 1-litre flasks are separately filled with the gases \(N_2\), Ne, \(N_2O\), and \(SO_3\) at the same temperature and pressure. The ratio of the total number of atoms of these gases present in different flasks would be:
- Given: All flasks have the same volume (1 L) at the same temperature and pressure.
- Step 1: At STP, 1 mole of any gas occupies 22.4 L. Therefore, 1 L contains \(\frac{1}{22.4}\) moles of gas.
- Step 2: Calculate the number of atoms for each gas.
- \(N_2\): 1 molecule contains 2 atoms.
\[
\text{Number of atoms} = \left(\frac{1}{22.4}\right) \times 2 \times 6.022 \times 10^{23}
\]
- Ne: 1 molecule contains 1 atom.
\[
\text{Number of atoms} = \left(\frac{1}{22.4}\right) \times 1 \times 6.022 \times 10^{23}
\]
- \(N_2O\): 1 molecule contains 3 atoms.
\[
\text{Number of atoms} = \left(\frac{1}{22.4}\right) \times 3 \times 6.022 \times 10^{23}
\]
- \(SO_3\): 1 molecule contains 4 atoms.
\[
\text{Number of atoms} = \left(\frac{1}{22.4}\right) \times 4 \times 6.022 \times 10^{23}
\]

- Step 3: The ratio of the number of atoms is:
\[
2 : 1 : 3 : 4
\]

- Answer: (3) \(2 : 1 : 3 : 4\)

---

#### A-6. The total number of g-molecules of \(SO_2Cl_2\) in 13.5 g of sulphuryl chloride is:
- Given: Mass of \(SO_2Cl_2\) = 13.5 g
- Step 1: Calculate the molar mass of \(SO_2Cl_2\).
\[
\text{Molar mass} = 32 + 2 \times 16 + 2 \times 35.5 = 32 + 32 + 71 = 135 \, \text{g/mol}
\]
- Step 2: Calculate the number of moles.
\[
\text{Number of moles} = \frac{13.5 \, \text{g}}{135 \, \text{g/mol}} = 0.1 \, \text{mol}
\]

- Answer: (1) 0.1

---

#### A-7. The number of sodium atoms in 2 moles of sodium ferrocyanide (\(Na_4[Fe(CN)_6]\)) is:
- Given: Formula = \(Na_4[Fe(CN)_6]\)
- Step 1: In one formula unit of \(Na_4[Fe(CN)_6]\), there are 4 sodium atoms.
- Step 2: For 2 moles of \(Na_4[Fe(CN)_6]\):
\[
\text{Number of sodium atoms} = 2 \, \text{mol} \times 4 \times 6.022 \times 10^{23} = 48 \times 10^{23}
\]

- Answer: (4) \(48 \times 10^{23}\)

---

#### A-8. 4.4 g of an unknown gas occupies 2.24 litres of volume at STP. The gas may be:
- Given: Mass = 4.4 g, Volume = 2.24 L at STP
- Step 1: At STP, 22.4 L corresponds to 1 mole. Therefore, 2.24 L corresponds to:
\[
\text{Moles} = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol}
\]
- Step 2: Calculate the molar mass.
\[
\text{Molar mass} = \frac{\text{Mass}}{\text{Moles}} = \frac{4.4 \, \text{g}}{0.1 \, \text{mol}} = 44 \, \text{g/mol}
\]
- Step 3: Identify the gas with a molar mass of 44 g/mol.
- \(N_2O\) (molar mass = 44 g/mol)
- \(CO_2\) (molar mass = 44 g/mol)

- Answer: (4) 1 & 3 Both

---

#### A-9. 5.6 litre of oxygen at STP contains:
- Given: Volume = 5.6 L at STP
- Step 1: At STP, 22.4 L corresponds to 1 mole. Therefore, 5.6 L corresponds to:
\[
\text{Moles} = \frac{5.6 \, \text{L}}{22.4 \, \text{L/mol}} = 0.25 \, \text{mol}
\]
- Step 2: Oxygen (\(O_2\)) is diatomic, so each mole contains \(2 \times 6.022 \times 10^{23}\) atoms.
\[
\text{Number of atoms} = 0.25 \, \text{mol} \times 2 \times 6.022 \times 10^{23} = 3.01 \times 10^{23}
\]

- Answer: (2) \(3.01 \times 10^{23}\) atoms

---

#### A-10. The volume of a gas in a discharge tube is \(1.12 \times 10^{-7}\) ml at STP. Then the number of molecules of gas in the tube is:
- Given: Volume = \(1.12 \times 10^{-7} \, \text{ml} = 1.12 \times 10^{-10} \, \text{L}\)
- Step 1: At STP, 22.4 L corresponds to \(6.022 \times 10^{23}\) molecules. Therefore:
\[
\text{Number of molecules} = \left(\frac{1.12 \times 10^{-10} \, \text{L}}{22.4 \, \text{L}}\right) \times 6.022 \times 10^{23} = 3.01 \times 10^{12}
\]

- Answer: (3) \(3.01 \times 10^{12}\)

---

#### A-11. A person adds 1.71 gram of sugar (\(C_{12}H_{22}O_{11}\)) in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar = 342):
- Given: Mass of sugar = 1.71 g, Molar mass of sugar = 342 g/mol
- Step 1: Calculate the number of moles of sugar.
\[
\text{Moles of sugar} = \frac{1.71 \, \text{g}}{342 \, \text{g/mol}} = 0.005 \, \text{mol}
\]
- Step 2: Each molecule of sugar (\(C_{12}H_{22}O_{11}\)) contains 12 carbon atoms.
\[
\text{Number of carbon atoms} = 0.005 \, \text{mol} \times 12 \times 6.022 \times 10^{23} = 3.6 \times 10^{22}
\]

- Answer: (1) \(3.6 \times 10^{22}\)

---

#### A-12. If \(V \, \text{ml}\) of the vapours of a substance at NTP weigh \(W \, \text{g}\). Then mol. w. of substance is:
- Given: Volume = \(V \, \text{ml} = V \times 10^{-3} \, \text{L}\), Weight = \(W \, \text{g}\)
- Step 1: At NTP, 22.4 L corresponds to 1 mole. Therefore, the number of moles in \(V \times 10^{-3} \, \text{L}\) is:
\[
\text{Moles} = \frac{V \times 10^{-3} \, \text{L}}{22.4 \, \text{L/mol}} = \frac{V}{22400} \, \text{mol}
\]
- Step 2: Molar mass = \(\frac{\text{Mass}}{\text{Moles}}\).
\[
\text{Molar mass} = \frac{W \, \text{g}}{\frac{V}{22400} \, \text{mol}} = \frac{W \times 22400}{V} \, \text{g/mol}
\]

- Answer: (1) \((W/V) \times 22400\)

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Section (B): Percentage composition and molecular formula

#### B-1. The number of atoms of Cr and O in a compound are \(4.8 \times 10^{10}\) and \(9.6 \times 10^{10}\) respectively. Its empirical formula is:
- Given: Number of Cr atoms = \(4.8 \times 10^{10}\), Number of O atoms = \(9.6 \times 10^{10}\)
- Step 1: Find the simplest whole number ratio of Cr to O.
\[
\text{Ratio of Cr to O} = \frac{4.8 \times 10^{10}}{4.8 \times 10^{10}} : \frac{9.6 \times 10^{10}}{4.8 \times 10^{10}} = 1 : 2
\]
- Step 2: The empirical formula is \(CrO_2\).

- Answer: (2) \(CrO_2\)

---

#### B-2. The empirical formula of a compound of molecular mass 120 is \(CH_2O\). The molecular formula of the compound is:
- Given: Empirical formula = \(CH_2O\), Molecular mass = 120
- Step 1: Calculate the empirical formula mass.
\[
\text{Empirical formula mass} = 12 + 2 \times 1 + 16 = 30
\]
- Step 2: Determine the multiplier \(n\).
\[
n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{120}{30} = 4
\]
- Step 3: Multiply the subscripts in the empirical formula by \(n\).
\[
\text{Molecular formula} = (CH_2O)_4 = C_4H_8O_4
\]

- Answer: (2) \(C_4H_8O_4\)

---

#### B-3. A hydrocarbon contains 75% of carbon. Then its molecular formula is:
- Given: Percentage of carbon = 75%, Percentage of hydrogen = 25%
- Step 1: Assume 100 g of the hydrocarbon.
- Mass of carbon = 75 g
- Mass of hydrogen = 25 g
- Step 2: Calculate the number of moles of each element.
\[
\text{Moles of C} = \frac{75 \, \text{g}}{12 \, \text{g/mol}} = 6.25 \, \text{mol}
\]
\[
\text{Moles of H} = \frac{25 \, \text{g}}{1 \, \text{g/mol}} = 25 \, \text{mol}
\]
- Step 3: Find the simplest whole number ratio.
\[
\text{Ratio of C to H} = \frac{6.25}{6.25} : \frac{25}{6.25} = 1 : 4
\]
- Step 4: The empirical formula is \(CH_4\).

- Answer: (1) \(CH_4\)

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Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{A-1: (2)} & \text{A-2: (1)} & \text{A-3: (3)} & \text{A-4: (2)} & \text{A-5: (3)} \\
\text{A-6: (1)} & \text{A-7: (4)} & \text{A-8: (4)} & \text{A-9: (2)} & \text{A-10: (3)} \\
\text{A-11: (1)} & \text{A-12: (1)} & \text{B-1: (2)} & \text{B-2: (2)} & \text{B-3: (1)}
\end{array}
}
\]