This chapter introduces fundamental concepts in chemistry, laying the groundwork for future studies. It covers the following key points:

**Matter and Its Properties**

**Definition of matter**: Anything that occupies space and has mass.

**States of matter: **Solid, liquid, and gas.

**Classification of matter: **Elements, compounds, and mixtures.

**Physical and chemical properties:** Properties that can be observed without changing the chemical composition and properties that involve a change in chemical composition, respectively.

**Measurement and Units**

**International System of Units (SI): **The standard system of units for measurement.

**Base units: **The fundamental units for measurement (e.g., meter, kilogram, second).

**Derived units:** Units derived from base units (e.g., volume, density).

**Significant figures:** The number of reliable digits in a measurement.

**Laws of Chemical Combination**

**Law of conservation of mass:** Mass is neither created nor destroyed in a chemical reaction.

**Atomic Theory**

**Dalton’s atomic theory:** A theory proposing that matter is composed of tiny, indivisible particles called atoms.

**Exercise **

**1. Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4**

**Ans : **

**1. H2O (water)**

Molar mass of H2O = (2 * 1.008 u) + (1 * 15.999 u) = 18.015 u

**2. CO2 (carbon dioxide)**

Molar mass of CO2 = (1 * 12.011 u) + (2 * 15.999 u) = 44.01 u

**3. CH4 (methane)**

Molar mass of CH4 = (1 * 12.011 u) + (4 * 1.008 u) = 16.043 u

**2. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).**

**Ans : **

**Molar mass of Na2SO4:**

- Atomic mass of sodium (Na) = 22.99 u
- Atomic mass of sulfur (S)

= 32.07 u

- Atomic mass of oxygen (O)

= 15.999 u

Molar mass of Na2SO4 = (2 * 22.99 u) + (1 * 32.07 u) + (4 * 15.999 u) = 142.04 u

**Mass percentage of each element:**

**Sodium (Na):**Mass of Na in Na2SO4 = 2 * 22.99 u = 45.98 u

Mass percentage of Na = (45.98 u / 142.04 u) * 100% ≈ 32.36%

**Sulfur (S):**Mass of S in Na2SO4 = 1 * 32.07 u = 32.07 u

Mass percentage of S = (32.07 u / 142.04 u) * 100% ≈ 22.56%

**Oxygen (O):**Mass of O in Na2SO4 = 4 * 15.999 u = 63.996 u

Mass percentage of O = (63.996 u / 142.04 u) * 100% ≈ 45.08%

**3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass**

**Ans :**

**1. Calculate the moles of each element:**

- Moles of iron = (69.9 g) / (55.845 g/mol)

= 1.25 mol

- Moles of oxygen = (30.1 g) / (15.999 g/mol) = 1.88 mol

**2. Find the simplest whole-number ratio of moles:**

- Mole ratio of iron to oxygen = 1.25 mol / 1.88 mol = 0.665
- To convert this to a whole-number ratio, we can multiply by a suitable factor. In this case, multiplying by 3 gives:
- Iron ratio = 0.665 * 3 ≈ 2
- Oxygen ratio = 1.88 * 3 ≈ 6

**3. Construct the empirical formula:**

The empirical formula is Fe2O6. However, this can be further simplified by dividing both subscripts by their greatest common divisor, which is 2.

**The empirical formula of the iron oxide is FeO3.**

**4. Calculate the amount of carbon dioxide that could be produced when**

** (i) 1 mole of carbon is burnt in air. **

**(ii) 1 mole of carbon is burnt in 16 g of dioxygen. **

**(iii) 2 moles of carbon are burnt in 16 g of dioxygen.**

**Ans : **

**(i) **

Assuming there is sufficient oxygen in the air, 1 mole of carbon will react with 1 mole of oxygen to produce 1 mole of carbon dioxide.

Therefore, 1 mole of carbon dioxide is **44.01 grams**.

**(ii) **

**Molar mass of O₂ = 2 * 16.00 g/mol = 32.00 g/mol**

16 grams of oxygen is equivalent to 16 g / 32.00 g/mol = 0.5 moles of oxygen.

Since the balanced equation shows a 1:1 ratio of carbon to oxygen, 0.5 moles of oxygen will react with 0.5 moles of carbon to produce 0.5 moles of carbon dioxide.

**Mass of CO₂ produced = 0.5 moles * 44.01 g/mol = 22.01 grams**

**(iii) **

As we found in (ii), 16 grams of oxygen is equivalent to 0.5 moles of oxygen. This time, we have 2 moles of carbon, but only enough oxygen to react with 0.5 moles of carbon.

Therefore, oxygen is the limiting reactant. The maximum amount of CO₂ that can be produced will be determined by the 0.5 moles of oxygen.

**Mass of CO₂ produced = 0.5 moles * 44.01 g/mol = 22.01 grams**

**5. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1**

**Ans : **

**Calculate the number of moles:**

- We have the volume of the solution (500 mL) and the molarity (0.375 M).
- 500 mL = 0.5 L
- Use the formula: moles = molarity × volume
- moles = 0.375 M × 0.5 L = 0.1875 moles

**Calculate the mass:**

- We have the moles and the molar mass (82.0245 g/mol).
- mass = moles × molar mass
- mass = 0.1875 moles × 82.0245 g/mol ≈ 15.38 grams

Therefore, you would need approximately 15.38 grams of sodium acetate to make 500 mL of a 0.375 M aqueous solution.

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%

**6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69%**

**Ans: **

Assume a 100g sample:

This simplifies calculations. If the sample is 100g, then 69g of it is nitric acid.

**Calculate the volume of the solution**:

Use the density formula: density = mass/volume

Rearrange to find volume: volume = mass/density

Volume = 100g / 1.41 g/mL ≈ 70.92 mL

Convert to liters: 70.92 mL = 0.07092 L

**Calculate the moles of nitric acid:**

Molar mass of HNO₃ = 1.01g/mol (H) + 14.01g/mol (N) + 16.00g/mol (O) = 63.02g/mol

Moles = mass/molar mass

Moles = 69g / 63.02g/mol ≈ 1.095 moles

**Calculate the concentration (molarity):**

Molarity = moles/volume

Molarity = 1.095 moles / 0.07092 L ≈ 15.44 mol/L

Therefore, the concentration of nitric acid in the sample is approximately 15.44 moles per liter.

**7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?**

**Ans : **

**Molar mass of CuSO₄:**

- Cu: 63.55 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol (4 atoms)
- Total: 159.61 g/mol

**Mass percentage of copper:**

- (Mass of copper / Molar mass of CuSO₄) * 100
- (63.55 g/mol / 159.61 g/mol) * 100 ≈ 39.81%

**Therefore, 39.81% of 100 g of CuSO₄ is copper. **

- 0.3981 * 100 g ≈ 39.81 g

**So, approximately 39.81 grams of copper can be obtained from 100 grams of copper sulfate.**

**8. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.**

**Ans : **

**1. Calculate the moles of each element:**

- Assume a 100g sample.
- Moles of iron = (69.9g) / (55.85 g/mol) ≈ 1.25 moles
- Moles of oxygen = (30.1g) / (16.00 g/mol) ≈ 1.88 moles

**2. Determine the simplest whole-number ratio of moles:**

- Divide both moles by the smaller value (1.25 moles):
- Iron: 1.25 moles / 1.25 moles = 1
- Oxygen: 1.88 moles / 1.25 moles ≈ 1.5

- Multiply both ratios by 2
- Iron: 1 * 2 = 2
- Oxygen: 1.5 * 2 = 3

**3. The empirical formula is Fe₂O₃.**

**Since the empirical formula and the molecular formula are the same in this case, the molecular formula of the iron oxide is also Fe₂O₃.**

**9. Calculate the atomic mass (average) of chlorine using the following data: % Natural Abundance**** **** **** **** ****Molar Mass Ans : **

** % Natural Abundance Molar Mass**

**35Cl 75.77 **** 34.9689 **

**37Cl 24.23 36.9659**

**Ans : **

**Isotope 1:**

Mass = 34.9689 u

Abundance = 75.77% = 0.7577

**Isotope 2:**

Mass = 36.9659 u

Abundance = 24.23% = 0.2423

**Substituting the values into the formula:**

Average atomic mass = (34.9689 u * 0.7577) + (36.9659 u * 0.2423)

≈ 26.468 u + 8.964 u

≈ 35.432 u

Therefore, the average atomic mass of chlorine is approximately 35.432 u.

**10. In three moles of ethane (C2H6), calculate the following: (i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms. (iii) Number of molecules of ethane**

**Ans : **

**(i) Number of moles of carbon atoms**

- Therefore, in 1 mole of ethane, there are 2 moles of carbon atoms.
- For 3 moles of ethane: 3 moles * 2 moles/mole =
**6 moles of carbon atoms**.

**(ii) Number of moles of hydrogen atoms**

- Therefore, in 1 mole of ethane, there are 6 moles of hydrogen atoms.
- For 3 moles of ethane: 3 moles * 6 moles/mole =
**18 moles of hydrogen atoms**.

**(iii) Number of molecules of ethane**

One mole of any substance contains 6.022 × 10²³ particles (Avogadro’s number).

- For 3 moles of ethane: 3 moles * 6.022 × 10²³ molecules/mole =
**1.8066 × 10²⁴ molecules of ethane**.

**11. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?**

**Ans : **

**Calculate the moles of sugar:**- Molar mass of C₁₂H₂₂O₁₁ = 342.3 g/mol
- Moles = mass / molar mass
- Moles = 20 g / 342.3 g/mol ≈ 0.0584 moles

**Calculate the concentration (molarity):**- Molarity = moles / volume
- Molarity = 0.0584 moles / 2 L ≈ 0.0292 mol/L

**Therefore, the concentration of sugar in the solution is approximately 0.0292 moles per liter.**

**12. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?**

**Ans : **

**Calculate the moles of methanol:**- Moles = molarity × volume
- Moles = 0.25 mol/L × 2.5 L = 0.625 mole

**Calculate the mass of methanol:**- Molar mass of methanol (CH₃OH) = 12.01g/mol (C) + 1.01g/mol (H) × 3 + 16.00g/mol (O) + 1.01g/mol (H) = 32.04 g/mol
- Mass = moles × molar mass
- Mass = 0.625 moles × 32.04 g/mol ≈ 20.03 g

**Use density to find the volume:**- Density = mass/volume
- Volume = mass/density
- Volume = 20.03 g / 0.793 kg/L = 20.03 g / 793 g/L ≈ 0.0253 L

**Convert the volume to milliliters:**

- 0.0253 L × 1000 mL/L = 25.3 mL

**Therefore, you would need approximately**

**13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1N m–2 If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal.**

**Ans : **

**Conversion Factors:**

- 1 g = 0.001 kg
- 1 cm = 0.01 m

**Calculations:**

**Convert mass to kilograms:**- 1034 g * (0.001 kg / 1 g) = 1.034 kg

**Convert area to square meters:**- 1 cm² * (0.01 m / 1 cm)² = 0.0001 m²

**Calculate pressure:**- Pressure = Force / Area
- Force (due to gravity)
- = mass * acceleration due to gravity (g)
- Assuming g ≈ 9.81 m/s²
- Pressure = (1.034 kg * 9.81 m/s²) / 0.0001 m²
- Pressure ≈ 101325 Pa

**Therefore, the pressure at sea level due to the mass of air is approximately 101,325 pascals.**

**14. What is the SI unit of mass? How is it defined?**

**Ans : **

The SI unit of mass is the **kilogram (kg)**. It is defined as the mass of a specific cylinder of platinum-iridium alloy kept at the International Bureau of Weights and Measures (BIPM) in Sèvres, France.

**15. Match the following prefixes with their multiples:**

** Prefixes**** **** Multiples **

** (i) micro 106**

** (ii) deca 109 **

** (iii) mega 10–6 **

** (iv) giga 10–15 **

** (v) femto 10**

**Ans : **

Prefix | Multiple |

(i) micro | 10⁻⁶ |

(ii) deca | 10 |

(iii) mega | 10⁶ |

(iv) giga | 10⁹ |

(v) femto | 10⁻¹⁵ |

**16. What do you mean by significant figures?**

**Ans : **

**Significant figures** are the digits in a number that are reliable or meaningful.

**17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass). **

**(i) Express this in per cent by mass. **

**(ii) Determine the molality of chloroform in the water sample.**

**Ans : **

**(i) Expressing the level of contamination in percent by mass**

- 1 ppm (parts per million) is equivalent to 1 part in 1,000,000 parts.

(15 / 1,000,000) * 100 = 0.0015%

**Therefore, the level of chloroform contamination is 0.0015% by mass.**

**(ii) Determining the molality of chloroform**

- Assuming the density of water is 1 g/mL, 1 L of water is equivalent to 1 kg of water.

**Calculate the moles of chloroform:**

- Molar mass of CHCl₃ = 12.01 g/mol (C) + 1.01 g/mol (H) + 35.45 g/mol (Cl) * 3 = 119.38 g/mol
- Moles of chloroform = (15 g / 1,000,000 g) / 119.38 g/mol ≈ 1.26 × 10⁻⁷ moles

**Calculate the molality:**

- Molality = moles of chloroform / mass of water (in kg)
- Molality ≈ 1.26 × 10⁻⁷ moles / 1 kg = 1.26 × 10⁻⁷ mol/kg

**18. Express the following in the scientific notation: **

**(i) 0.0048 **

**(ii) 234,000 **

**(iii) 8008 **

**(iv) 500.0 **

**(v) 6.0012**

**Ans : **

(i) 0.0048 = 4.8 × 10⁻³

(ii) 234,000 = 2.34 × 10⁵

(iii) 8008 = 8.008 × 10³

(iv) 500.0 = 5.000 × 10²

(v) 6.0012 = 6.0012 × 10⁰

**19. How many significant figures are present in the following? **

**(i) 0.0025 **

**(ii) 208 **

**(iii) 5005**

**(iv) 126,000 **

**(v) 500.0 **

**(vi) 2.0034**

**Ans : **

(i) **0.0025:** There are **2** significant figures (2 and 5). The leading zeros are not significant.

(ii) **208:** There are **3** significant figures (2, 0, and 8).

(iii) **5005:** There are **4** significant figures (5, 0, 0, and 5).

(iv) **126,000:** There are **3** significant figures (1, 2, and 6). The trailing zeros are not significant unless there’s a decimal point.

(v) **500.0:** There are **4** significant figures (5, 0, 0, and 0). The trailing zeros after the decimal point are significant.

(vi) **2.0034:** There are **5** significant figures (2, 0, 0, 3, and 4)

**20. Round up the following upto three significant figures: **

**(i) 34.216 **

**(ii) 10.4107 **

**(iii) 0.04597 **

**(iv) 2808**

**Ans : **

(i) 34.216 rounded to three significant figures is **34.2**.

(ii) 10.4107 rounded to three significant figures is **10.4**.

(iii) 0.04597 rounded to three significant figures is **0.046**.

(iv) 2808 rounded to three significant figures is **2810**.

**21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:**

**Mass of dinitrogen Mass of dioxygen **

**(i) 14 g 16 g **

**(ii) 14 g 32 g **

**(iii) 28 g 32 g **

**(iv) 28 g 80 g**

**(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. **

**(b) Fill in the blanks in the following conversions: **

**(i) 1 km = …………………. mm = …………………. pm **

**(ii) 1 mg = …………………. kg = …………………. ng **

**(iii) 1 mL = …………………. L = …………………. dm3**

**Ans **

It is a Law of Multiple Proportions. This law states:

When two elements combine to form more than one compound, the ratio of the masses of one element that combine with a fixed mass of the other element is a simple whole number ratio

**(b) Filling in the Blanks**

(i) 1 km = **1,000,000** mm = **1,000,000,000,000** pm

(ii) 1 mg = **0.001** kg = **1,000,000** ng

(iii) 1 mL = **0.001** L = **0.001** dm³

**22. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns**

**Ans : **

To calculate the distance covered by light in 2.00 ns, we can use the formula:

distance = speed × time

where:

- speed of light = 3.0 × 10⁸ m/s
- time = 2.00 ns = 2.00 × 10⁻⁹ s

distance = (3.0 × 10⁸ m/s) × (2.00 × 10⁻⁹ s)

distance = 6.00 × 10⁻¹ m

Therefore, the distance covered by light in 2.00 ns is **0.600 meters**.

**23. In a reaction A + B2 AB2 Identify the limiting reagent, if any, in the following reaction mixtures. **

**(i) 300 atoms of A + 200 molecules of B **

**(ii) 2 mol A + 3 mol B **

**(iii) 100 atoms of A + 100 molecules of B **

**(iv) 5 mol A + 2.5 mol B **

**(v) 2.5 mol A + 5 mol B**

**Ans : **

(i) 300 atoms of A + 200 molecules of B

- 200 molecules of B₂ will react with 200 atoms of A.
- This will leave 100 atoms of A unreacted.

Therefore, **B₂ is the limiting reagent** in this case.

(ii) 2 mol A + 3 mol B

- 2 moles of A will react with 2 moles of B₂.
- This will leave 1 mole of B₂ unreacted.

Therefore, **A is the limiting reagent** in this case.

(iii) 100 atoms of A + 100 molecules of B

- 100 atoms of A will react with 100 molecules of B₂ completely.

Therefore, **neither A nor B₂ is the limiting reagent** in this case. They are present in stoichiometric amounts.

(iv) 5 mol A + 2.5 mol B

- 2.5 moles of B₂ will react with 2.5 moles of A.
- This will leave 2.5 moles of A unreacted.

Therefore, **B₂ is the limiting reagent** in this case.

(v) 2.5 mol A + 5 mol B

- 2.5 moles of A will react with 2.5 moles of B₂.
- This will leave 2.5 moles of B₂ unreacted.

Therefore, **A is the limiting reagent** in this case.

**24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g) 2NH3 (g) **

**(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. **

**(ii) Will any of the two reactants remain unreacted? **

**(iii) If yes, which one and what would be its mass?**

**Ans : **

(i) **Mass of ammonia produced:** 5666.67 grams

(ii) **Yes**, one reactant will remain unreacted.

(iii) **Dinitrogen (N₂) will remain unreacted.**

**25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?**

**Ans : **

0.50 mol Na₂CO₃ refers to a specific quantity of the compound.

0.50 M Na₂CO₃ refers to the concentration of Na₂CO₃ in a solution.

**26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced? **

**Ans : **

The balanced chemical equation for the reaction is:

2H₂(g) + O₂(g) → 2H₂O(g)

10 volumes of hydrogen gas would react with 5 volumes of oxygen gas to produce 10 volumes of water vapor.

**27. Convert the following into basic units: **

**(i) 28.7 pm **

**(ii) 15.15 pm **

**(iii) 25365 m**

**Ans : **

(i) 28.7 pm

- 1 picometer (pm) = 10⁻¹² meters
- Therefore, 28.7 pm = 28.7 × 10⁻¹² m = 2.87 × 10⁻¹¹ m

(ii) 15.15 pm

- Using the same conversion factor:
- 15.15 pm = 15.15 × 10⁻¹² m = 1.515 × 10⁻¹¹ m

(iii) 25365 m

- Meters are already the basic unit of length, so no conversion is needed.
- 25365 m = 25365 m

**28. Which one of the following will have the largest number of atoms? **

**(i) 1 g Au (s) **

**(ii) 1 g Na (s) **

**(iii) 1 g Li (s) (iv) 1 g of Cl2(g)**

**Ans : **

**Molar masses:**

- Au: 197 g/mol
- Na: 23 g/mol
- Li: 7 g/mol
- Cl₂: 71 g/mol

**Calculating moles:**

(i) Moles of Au = 1 g / 197 g/mol ≈ 0.0051 moles

(ii) Moles of Na = 1 g / 23 g/mol

≈ 0.0435 moles

(iii) Moles of Li = 1 g / 7 g/mol ≈ 0.1429 moles

(iv) Moles of Cl₂ = 1 g / 71 g/mol ≈ 0.0141 moles

**Calculating number of atoms:**

- Number of atoms = moles × Avogadro’s number (6.022 × 10²³)

(i) Number of Au atoms ≈ 0.0051 moles × 6.022 × 10²³ atoms/mol ≈ 3.07 × 10²¹ atoms

(ii) Number of Na atoms ≈ 0.0435 moles × 6.022 × 10²³ atoms/mol ≈ 2.62 × 10²² atoms

(iii) Number of Li atoms ≈ 0.1429 moles × 6.022 × 10²³ atoms/mol ≈ 8.60 × 10²² atoms

(iv) Number of Cl atoms ≈ 0.0141 moles × 6.022 × 10²³ atoms/mol ≈ 8.49 × 10²¹ atoms

**Comparing the number of atoms:**

- Li has the largest number of atoms (8.60 × 10²² atoms).

**29. Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).**

**Ans : **

**Assume 1 liter of solution:**Since the density of water is assumed to be 1, 1 liter of water is approximately equal to 1 kg of water.**Calculate moles of water:**- Molar mass of water (H₂O) = 18 g/mol
- Moles of water = mass / molar mass = 1000 g / 18 g/mol ≈ 55.56 moles

**Use mole fraction to find moles of ethanol:**- Mole fraction of ethanol = moles of ethanol / total moles
- 0.040 = moles of ethanol / (moles of ethanol + 55.56 moles)

**Solve for moles of ethanol:**- 0.040 * (moles of ethanol + 55.56 moles) = moles of ethanol
- 2.2224 + 0.040 * moles of ethanol = moles of ethanol
- 2.2224 = 0.96 * moles of ethanol
- moles of ethanol ≈ 2.31 moles

**Calculate molarity:**- Molarity = moles of ethanol / volume of solution
- Molarity = 2.31 moles / 1 L = 2.31 M

**30. What will be the mass of one 12C atom in g?**

**Ans : **

The mass of one ¹²C atom is approximately 1.993 × 10⁻²³ grams.

**31 . How many significant figures should be present in the answer of the following calculations? **

**(i) 0.02856 298.15 × × 0.112 / 0.5785 **

**(ii) 5 × 5.364 **

**(iii) 0.0125 + 0.7864 + 0.0215**

**Ans : **

**(i) 0.02856 * 298.15 * 0.112 / 0.5785**

- 0.02856 has 4 significant figures.
- 298.15 has 6 significant figures.
- 0.112 has 3 significant figures.
- 0.5785 has 4 significant figures.

The factor with the least significant figures is 0.112, so the result should have **3 significant figures**.

**(ii) 5 * 5.364**

- 5 is considered an exact number and has an unlimited number of significant figures.
- 5.364 has 4 significant figures.

Therefore, the result should have **4 significant figures**.

**(iii) 0.0125 + 0.7864 + 0.0215**

- 0.0125 has 3 decimal places.
- 0.7864 has 4 decimal places.
- 0.0215 has 3 decimal places.

**3 significant figures**.

**32. Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:**

**Ans : **

The formula for calculating the molar mass is:

Molar mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + …

Using the given data, we can calculate the molar mass of argon as follows:

Molar mass = (35.96755 g/mol * 0.337%) + (37.96272 g/mol * 0.063%) + (39.9624 g/mol * 99.600%)

Molar mass = (35.96755 g/mol * 0.00337) + (37.96272 g/mol * 0.00063) + (39.9624 g/mol * 0.99600)

Molar mass ≈ 0.1215 g/mol + 0.0239 g/mol + 39.7624 g/mol

Molar mass ≈ 39.9078 g/mol

**Therefore, the molar mass of naturally occurring argon is approximately 39.9078 g/mol.**

**33. Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.**

**Ans : **

**Case (i): 52 moles of Ar**

- Number of atoms
- = moles × Avogadro’s number
- Number of atoms = 52 moles × 6.022 × 10²³ atoms/mol ≈ 3.13 × 10²⁵ atoms

**Case (ii): 52 u of He**

- 52 u is the mass of a single helium atom.
- Since 1 mole of helium contains 6.022 × 10²³ atoms, 1 atom of helium is equal to 1/6.022 × 10²³ moles.
- Number of atoms
- = moles × Avogadro’s number
- Number of atoms = (52 u / 4 u/mol) × 6.022 × 10²³ atoms/mol ≈ 7.83 × 10²³ atoms

**Case (iii): 52 g of He**

- First, calculate the moles of helium:
- Moles of He = mass / molar mass = 52 g / 4 g/mol = 13 moles

- Then, calculate the number of atoms:
- Number of atoms
- = moles × Avogadro’s number
- Number of atoms = 13 moles × 6.022 × 10²³ atoms/mol ≈ 7.83 × 10²⁴ atoms

**34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.**

**Ans : **

**(i) Empirical Formula**

**Step 1: **

Moles of CO₂ = mass / molar mass = 3.38 g / 44.01 g/mol ≈ 0.0768 moles

Moles of H₂O = mass / molar mass = 0.690 g / 18.02 g/mol ≈ 0.0383 moles

**Step 2: **

From CO₂, moles of C = moles of CO₂ = 0.0768 moles

From H₂O, moles of H = 2 * moles of H₂O = 2 * 0.0383 moles = 0.0766 moles

**Step 3: Find the simplest whole-number ratio of C and H**

Ratio of C to H = 0.0768 moles / 0.0766 moles ≈ 1.003

This is approximately a 1:1 ratio.

Therefore, the empirical formula is CH.

**(ii) Molar Mass of the Gas**

Given, 10.0 L of the gas at STP weighs 11.6 g

At STP, 1 mole of any gas occupies 22.4 L.

So, 10.0 L of the gas is equivalent to 10.0 L / 22.4 L/mol = 0.446 moles.

Molar mass = mass / moles = 11.6 g / 0.446 moles ≈ 26 g/mol

**(iii) Molecular Formula**

To find the molecular formula, we need to compare the molar mass of the empirical formula (CH) with the calculated molar mass.

Molar mass of CH = 12.01 g/mol (C) + 1.01 g/mol (H) = 13.02 g/mol

Ratio of calculated molar mass to empirical formula molar mass = 26 g/mol / 13.02 g/mol ≈ 2

Molecular formula = (CH)₂ = C₂H₂

**Therefore, the welding gas is acetylene (C₂H₂). **

**35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?**

**Ans : **

**Calculate moles of HCl:**- Moles = Molarity × Volume (in liters)
- Moles of HCl = 0.75 mol/L × 0.025 L = 0.01875 moles HCl.
- So, moles of CaCO₃ = 0.01875 moles HCl / 2 = 0.009375 moles CaCO₃

**Calculate mass of CaCO₃:**- Molar mass of CaCO₃ = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3 * 16.00 g/mol (O) = 100.09 g/mol
- Mass of CaCO₃ = moles × molar mass = 0.009375 moles * 100.09 g/mol ≈ 0.938 g

**Therefore, 0.938 grams of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl.**

**36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide?**

**Ans : **

**Calculate moles of MnO₂:**- Molar mass of MnO₂ = 54.94 g/mol (Mn) + 2 * 16.00 g/mol (O) = 86.94 g/mol
- Moles of MnO₂ = mass / molar mass = 5.0 g / 86.94 g/mol ≈ 0.0575 moles

**Determine moles of HCl needed:**- From the balanced equation, 4 moles of HCl react with 1 mole of MnO₂.
- So, moles of HCl = 4 * moles of MnO₂ = 4 * 0.0575 moles ≈ 0.230 moles

**Calculate mass of HCl:**- Molar mass of HCl = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
- Mass of HCl = moles * molar mass = 0.230 moles * 36.46 g/mol ≈ 8.38 g