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
