Exercise 5(A)
1. State :
(a) Gay-Lussac’s Law of combining volumes, (2011)
(b) Avogadro’s law.
Ans: (a) Gay-Lussac’s Law of Combining Volumes
Gay-Lussac’s Law of Combining Volumes states that when gases react with each other, they do so in volumes which bear a simple whole number ratio to one another and to the volumes of the gaseous products, provided all volumes are measured at the same temperature and pressure.
In simpler terms: If you measure the volumes of reacting gases and the volumes of any gaseous products, you will find that these volumes can be expressed as small whole numbers. For example, in the reaction where hydrogen combines with oxygen to form water vapour, two volumes of hydrogen always combine with one volume of oxygen to produce two volumes of steam.
(b) Avogadro’s Law
Avogadro’s Law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.
In simpler terms: It doesn’t matter what type of gas it is—one litre of hydrogen, one litre of oxygen, or one litre of any other gas—if they are at the same temperature and pressure, they will all have the exact same number of molecules. This law was crucial in distinguishing between atoms and molecules and provided the foundation for understanding gaseous reactions.
2.(a) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.
(b) Differentiate between N2 and 2N.
Ans: (a) Atomicity refers to the number of atoms present in a single molecule of an element. It classifies elements based on whether their molecules are composed of single atoms, two atoms, or more. For the given elements, Hydrogen exists as a diatomic molecule (H₂), so its atomicity is 2. Phosphorus commonly exists as a tetra-atomic molecule (P₄) in its white form, so its atomicity is 4. Sulphur, in its most stable form at room temperature, exists as an octa-atomic molecule (S₈), which means its atomicity is 8.
(b) The difference between N₂ and 2N lies in their fundamental chemical meaning. N₂ represents a molecule of nitrogen gas, which is the stable form of the element in nature. This molecule is composed of two nitrogen atoms chemically bonded together. In contrast, 2N does not represent a molecule; it signifies two separate, individual atoms of nitrogen that are not bonded to each other. It is often used in chemical equations to denote the atomic state or to show that two nitrogen atoms are required for a reaction.
3. Explain Why ?
(a) “The number of atoms in a certain volume of hydrogen is twice the number of atoms in the same volume of helium at the same temperature and pressure.”
(b) “When stating the volume of a gas, the pressure and temperature should also be given.”
(c) Inflating a balloon seems to violate Boyle’s law.
Ans:(a) This occurs because hydrogen (H₂) and helium (He) are both diatomic and monatomic gases respectively under standard conditions. According to Avogadro’s law, equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. Therefore, a certain volume will contain the same number of molecules for both gases. However, each hydrogen molecule (H₂) is composed of two atoms, while each helium molecule is a single atom. Consequently, the total number of atoms in the hydrogen gas is twice that in the same volume of helium.
(b) The volume of a gas is highly dependent on its pressure and temperature because gases are compressible and expand when heated. Stating a volume without specifying the pressure and temperature is ambiguous, as that same amount of gas can occupy vastly different volumes under different conditions. Therefore, providing the pressure and temperature is essential to define the state of the gas and make the volume measurement meaningful and reproducible.
(c) Inflating a balloon seems to violate Boyle’s law because the law states that at a constant temperature, the volume of a gas is inversely proportional to its pressure. When you inflate a balloon, you are adding more air molecules, increasing the internal pressure, yet the volume also increases. This seems contradictory. The resolution is that Boyle’s law applies to a fixed mass of gas. When inflating, the mass of gas inside the balloon is not constant; it is increasing. The increase in volume is due to this addition of more air, not because the existing air is expanding against its pressure.
NUMERICAL PROBLEMS
4.(a) Calculate the volume of oxygen at STP required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.
2CO+O2→2CO2
Ans:Based on the balanced chemical equation 2CO + O₂ → 2CO₂, the reaction shows that 2 volumes of carbon monoxide need 1 volume of oxygen. Therefore, for 100 litres of carbon monoxide, the required oxygen volume is 50 litres at standard temperature and pressure.
(b) 200 cm3 of hydrogen and 150 cm3 of oxygen are mixed and ignited, as per following reaction,
2H2+O2→2H2O
What volume of oxygen remains unreacted ?
Ans: In the reaction 2H₂ + O₂ → 2H₂O, the gases react in a 2:1 volume ratio.
We have 200 cm³ of H₂, which would require 100 cm³ of O₂ to react completely (since 200:100 = 2:1).
We started with 150 cm³ of O₂, so after 100 cm³ reacts, the remaining oxygen is:
150 cm³ – 100 cm³ = 50 cm³.
Thus, 50 cm³ of oxygen remains unreacted.
5. 24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling, the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
Ans: This experiment supports the Law of Combining Volumes (or Gay-Lussac’s Law of Gaseous Volumes).
The initial volumes are:
Marsh gas (CH₄) = 24 cc
Oxygen = 106 cc
After explosion and cooling, the mixture volume is 82 cc, which includes 58 cc of unreacted oxygen.
Volume of oxygen used in reaction = Initial oxygen − Unchanged oxygen
= 106 cc − 58 cc = 48 cc oxygen used.
Let the reaction be:
CH₄ + 2O₂ → CO₂ + 2H₂O (liquid after cooling)
From stoichiometry: 1 vol CH₄ reacts with 2 vol O₂.
So, 24 cc CH₄ would require 2 × 24 = 48 cc O₂, which matches exactly the oxygen consumed.
The products after cooling:
CO₂ volume produced = volume of CH₄ used (from stoichiometry) = 24 cc
Water is liquid, so negligible volume.
Final gaseous mixture = CO₂ (24 cc) + Unused O₂ (58 cc) = 82 cc, matching the observed volume.
Since the volume ratio CH₄ : O₂ : CO₂ = 24 : 48 : 24 = 1 : 2 : 1 follows small whole numbers, it confirms the Law of Combining Volumes.
6. What volume of oxygen would be required to burn 400 mL of acetylene [C2H2] ? Also, calculate the volume of carbon dioxide formed.
2C2H2+5O2→4CO2+2H2O(l)
Ans: According to the balanced equation 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O(l), 2 volumes of acetylene require 5 volumes of oxygen.
For 400 mL of acetylene, the required oxygen volume is calculated as (5/2) × 400 mL = 1000 mL.
Similarly, 2 volumes of acetylene produce 4 volumes of carbon dioxide, so the volume of CO₂ formed is (4/2) × 400 mL = 800 mL.Thus, burning 400 mL of acetylene requires 1000 mL of oxygen and produces 800 mL of carbon dioxide.
7. 112 cm3 of H2S(g) is mixed with 120 cm3 of Cl2(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate
(i) the volume of gaseous product formed
(ii) composition of the resulting mixture.
Ans: The balanced chemical equation for the reaction is:
H2S(g)+Cl2(g)→2HCl(g)+S(s)
At STP, 112 cm³ of H₂S is 0.005 mol (112/22400) and 120 cm³ of Cl₂ is approximately 0.00536 mol. From the stoichiometry, 1 mole of H₂S reacts with 1 mole of Cl₂. Since H₂S has the lesser amount, it is the limiting reactant.
(i) Volume of gaseous product formed:
The only gaseous product is HCl. 0.005 mol of H₂S produces 0.01 mol of HCl. At STP, this is
0.01×22400=224 cm3
(ii) Composition of the resulting mixture:
The reaction consumes all 0.005 mol of H₂S and an equal amount of Cl₂. The leftover Cl₂ is
0.00536−0.005=0.00036 mol, which is 0.00036×22400≈8 cm3
The solid sulfur formed is 0.005 mol and is part of the mixture.
Thus, the final mixture consists of 224 cm³ of HCl gas, 8 cm³ of unreacted Cl₂ gas, and 0.005 mol of solid sulfur.
8. 2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed, after writing a balanced equation :
Ethane + Oxygen → Carbon dioxide + Water vapour
Ans: In the balanced equation for the combustion of ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, 2 volumes of ethane require 7 volumes of oxygen.
Given 600 cc of ethane, the required oxygen is (7/2) × 600 = 2100 cc.
Since 2500 cc of oxygen was supplied, the unused oxygen is 2500 – 2100 = 400 cc.
From the equation, 2 volumes of ethane produce 4 volumes of carbon dioxide.
So, 600 cc of ethane will produce (4/2) × 600 = 1200 cc of carbon dioxide.
Final Answer: Unused oxygen = 400 cc, Carbon dioxide formed = 1200 cc.
9. What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273°C and 380 mm of Hg pressure?
C2H4+3O2→2CO2+2H2O
Ans: In the balanced equation for the combustion of ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, 2 volumes of ethane require 7 volumes of oxygen.
Given 600 cc of ethane, the required oxygen is (7/2) × 600 = 2100 cc.
Since 2500 cc of oxygen was supplied, the unused oxygen is 2500 – 2100 = 400 cc.
From the equation, 2 volumes of ethane produce 4 volumes of carbon dioxide.
So, 600 cc of ethane will produce (4/2) × 600 = 1200 cc of carbon dioxide.
Final Answer: Unused oxygen = 400 cc, Carbon dioxide formed = 1200 cc.
10. Calculate the volume of HCl gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at STP.
CH4+2Cl2→CH2Cl2+2HCl
Ans: When 40 mL of methane (CH₄) reacts completely with chlorine, the balanced equation is:
CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl
According to Gay-Lussac’s law of combining volumes, the mole ratio is the same as the volume ratio for gases at STP.
From the equation:
1 volume CH₄ produces 2 volumes HCl
So, 40 mL CH₄ produces:
Volume of HCl = 40 × 2 = 80 mL
Also, 1 volume CH₄ reacts with 2 volumes Cl₂
So, 40 mL CH₄ requires:
Volume of Cl₂ = 40 × 2 = 80 mL
Final answer:
Volume of HCl gas formed = 80 mL
Volume of chlorine gas required = 80 mL
11. What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions ? (Assuming oxygen is 1/5th of air)
C3H8+5O2→3CO2+4H2O
Ans: Based on the balanced chemical equation, 1 volume of propane requires 5 volumes of oxygen for complete combustion. The 500 cm³ sample of air contains 100 cm³ of oxygen, since oxygen constitutes one-fifth of the air’s volume. Applying the 1:5 ratio, the volume of propane that can be burnt is calculated as 100 cm³ divided by 5, which equals 20 cm³.
12. 450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of the resulting mixture.
2NO+O2→2NO2
Ans: When 450 cm³ of nitrogen monoxide (NO) and 200 cm³ of oxygen (O₂) are mixed, the reaction 2NO + O₂ → 2NO₂ occurs.From the stoichiometry, 2 volumes of NO react with 1 volume of O₂.
Thus, 450 cm³ of NO would require 225 cm³ of O₂.
However, only 200 cm³ of O₂ is available, so O₂ is the limiting reactant.
200 cm³ of O₂ will react with 400 cm³ of NO (since 2NO : 1O₂).
This produces 400 cm³ of NO₂ (since 2NO₂ : 2NO).
After reaction, NO left = 450 − 400 = 50 cm³, and O₂ left = 0 cm³.
Resulting mixture: 400 cm³ NO₂ and 50 cm³ NO.
13. If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas. (2015)
Ans: When 6 litres of hydrogen and 4 litres of chlorine are mixed, they react to form hydrogen chloride gas according to the reaction:
H₂ + Cl₂ → 2HCl.
Here, hydrogen and chlorine react in a 1:1 volume ratio. Since only 4 litres of chlorine is available, it is the limiting reactant.
Thus, 4 litres of hydrogen reacts with 4 litres of chlorine to produce 8 litres of HCl.
This leaves (6 − 4) = 2 litres of hydrogen unreacted.
When water is added, HCl gas dissolves completely in water, but hydrogen gas does not.
Therefore, the residual gas is the unreacted hydrogen, and its volume is 2 litres.
14. Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.
4NH3+5O2→4NO+6H2O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure ?
Ans: When 27 litres of the reactants (ammonia and oxygen) are consumed in the given reaction:
4NH3+5O2→4NO+6H2O
the total mole ratio of gaseous reactants is
4+5=9 moles.
Under the same temperature and pressure, volume is proportional to moles.
The products include nitrogen monoxide (NO) and water, but since water is liquid under standard conditions, only NO is considered as a gas here.
From the equation, 9 moles of reactants produce 4 moles of NO.
Thus, volume of NO produced =
49×2794 ×27 litres = 12 litres.
Final answer: 12 litres of nitrogen monoxide is produced.
15. A mixture of hydrogen and chlorine occupying 36 cm3 exploded. On shaking it with water, 4 cm3 of hydrogen was left behind. Find the composition of the mixture.
Ans: When 36 cm³ of hydrogen and chlorine mixture exploded, the reaction was:
H₂ + Cl₂ → 2HCl.
The product HCl dissolved in water on shaking, leaving 4 cm³ of unreacted hydrogen.
This means the hydrogen was in excess.
Let the volume of chlorine be x cm³.
Then hydrogen used in the reaction = x cm³ (since 1:1 mole ratio).
Total hydrogen present = hydrogen used + leftover hydrogen = x+4 cm³.
Total mixture volume = hydrogen present + chlorine present
(x+4)+x=36
2x+4=36
2x=32
x=16
So, chlorine = 16 cm³, hydrogen = 16+4=20 cm³.
Composition: Hydrogen = 20 cm³, Chlorine = 16 cm³.
16. What volume of air (containing 20% O2by volume) will be required to burn completely 10 cm3 each of methane and acetylene.
CH4+2O2→CO2+2H2O
2C2H2+5O2→4CO2+2H2O
Ans: To burn 10 cm³ of methane (CH₄), the reaction CH₄ + 2O₂ → CO₂ + 2H₂O shows that 1 volume of CH₄ requires 2 volumes of O₂. Thus, 10 cm³ CH₄ requires 20 cm³ of O₂.
For 10 cm³ of acetylene (C₂H₂), the reaction 2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O indicates that 2 volumes of C₂H₂ require 5 volumes of O₂. So, 10 cm³ C₂H₂ requires (5/2) × 10 = 25 cm³ of O₂.
Total O₂ needed = 20 + 25 = 45 cm³.
Given that air contains 20% O₂ by volume, the volume of air required = (100 / 20) × 45 = 225 cm³.
17. LPG has 60% propane and 40% butane : 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.
Ans: C3H8+5O2→3CO2+4H2O
2C4H10+13O2→8CO2+10H2O
Ans: We are given that LPG contains 60% propane (C₃H₈) and 40% butane (C₄H₁₀) by volume, and 10 litres of the mixture is burnt.
From the combustion equations:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 volume of propane produces 3 volumes of CO₂.
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
2 volumes of butane produce 8 volumes of CO₂, so 1 volume of butane produces 4 volumes of CO₂.
Step 1: Volumes of each gas in 10 litres
Propane = 60% of 10 L = 6 L
Butane = 40% of 10 L = 4 L
Step 2: CO₂ from propane
6 L C₃H₈ → 6 × 3 = 18 L CO₂
Step 3: CO₂ from butane
4 L C₄H₁₀ → 4 × 4 = 16 L CO₂
Step 4: Total CO₂
18 L + 16 L = 34 L
Thus, the volume of carbon dioxide added to the atmosphere is 34 litres.
18.200cm3 of CO2 is collected at STP when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in the original mixture.
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
Ans: In the given reaction, 2 volumes of C₂H₂ react with 5 volumes of O₂ to produce 4 volumes of CO₂.
We know 18.200 cm³ of CO₂ is collected at STP.
From the stoichiometry, 4 volumes of CO₂ are produced from 2 volumes of C₂H₂.
So, volume of C₂H₂ used = (2/4) × 18.200 = 9.100 cm³.
Similarly, 4 volumes of CO₂ come from 5 volumes of O₂.
So, volume of O₂ used = (5/4) × 18.200 = 22.750 cm³.
Thus, the original mixture at STP contained 9.100 cm³ of acetylene and 22.750 cm³ of oxygen.
19. You have collected (a) 2 litres of CO2
(b) 3 litres of chlorine (c) 5 litres of hydrogen (d) 4 litres of nitrogen and (e) 1 litre of SO2, under similar conditions of temperature and pressure. Which gas sample will have :
(a) the greatest number of molecules, and
(b) the least number of molecules ?
Justify your answers.
Ans: Under the same temperature and pressure, equal volumes of all gases contain an equal number of molecules (Avogadro’s law). Therefore, the number of molecules in each gas sample depends only on its volume.
The gas with the largest volume of 5 litres (hydrogen) will have the greatest number of molecules. Conversely, the gas with the smallest volume of 1 litre (sulfur dioxide, SO₂) will have the least number of molecules. The chemical identity of the gas does not affect the number of molecules, only the volume does under these conditions.
20. The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.
Ans: Under the same temperature and pressure, equal volumes of gases contain an equal number of molecules (Avogadro’s law).
The volume of nitrogen is 20 L and it contains *x* molecules.
Chlorine (Volume: 10 L):
Since 20 L contains *x* molecules, 10 L will contain half the number of molecules.
→ Number of molecules = x/2
Ammonia (Volume: 40 L):
40 L is double the volume of nitrogen, so it will contain double the molecules.
→ Number of molecules = 2x
Sulphur dioxide (Volume: 5 L):
5 L is one-fourth of 20 L, so it will contain one-fourth the molecules.
→ Number of molecules = x/4
21. If 100 cm3 of oxygen contains Y molecules, how many molecules of nitrogen will be present in 50 cm3 of nitrogen under the same conditions of temperature and pressure?
Ans: Under the same temperature and pressure, an equal volume of any gas contains the same number of molecules.Since 100 cm³ of oxygen contains Y molecules, the number of molecules in 50 cm³ of any gas (including nitrogen) will be half of Y.Therefore, the number of nitrogen molecules present in 50 cm³ is Y/2.
Exercise 5(B)
1.(a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.
(b) What is the value of Avogadro’s number?
(c) What is the value of molar volume of a gas at S.T.P.?
Ans: (a) The relative atomic mass of chlorine being 35.5 a.m.u. indicates that a natural sample of chlorine atoms is a mixture of isotopes, primarily chlorine-35 and chlorine-37. The value 35.5 is not the mass of a single atom, but rather the average mass of all the atoms in the sample, calculated by considering the proportional abundance of each isotope.
(b) Avogadro’s number is defined as the number of constituent particles (atoms, molecules, or ions) present in one mole of a substance. Its fixed value is 6.022 × 10²³ mol⁻¹.
(c) The molar volume of any gas at Standard Temperature and Pressure (S.T.P.), which is defined as 0°C (273 K) and 1 atmosphere of pressure, occupies a volume of 22.4 litres or 22.4 dm³.
2. Define or explain the terms :
(a) vapour density,
(b) molar volume,
(c) relative atomic mass,
(d) relative molecular mass,
(e) Avogadro’s number,
(f) Gram atom,
(g) Mole.
Ans: (a) Vapour Density
Vapour density is defined as the mass of a certain volume of a gas or vapour compared to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure. It is a unitless quantity. Practically, it provides a simple way to determine the molecular mass of a gas, as the molecular mass is approximately twice its vapour density.
(b) Molar Volume
Molar volume is the volume occupied by one mole of any substance in the gaseous state at a specific temperature and pressure. At standard temperature and pressure (STP, 0°C and 1 atm), one mole of any ideal gas occupies a volume of approximately 22.4 litres. This concept is fundamental in relating the volume of a gas to the amount of substance present.
(c) Relative Atomic Mass
Relative atomic mass is the average mass of an atom of a chemical element, compared to one-twelfth the mass of a carbon-12 atom.It is the standard atomic weight value found on the periodic table and represents the average mass of all the naturally occurring isotopes of an element.
(d) Relative Molecular Mass
Relative molecular mass is the sum of the relative atomic masses of all the atoms that constitute a molecule of a substance. Like relative atomic mass, it is a dimensionless number. It was formerly known as molecular weight and is calculated directly from the chemical formula and the standard atomic weights.
(e) Avogadro’s Number
Avogadro’s number is the fundamental constant that represents the number of constituent particles (atoms, molecules, or ions) present in one mole of a substance. Its value is approximately 6.022 × 10²³. This number provides a crucial link between the macroscopic world we can measure and the microscopic world of atoms and molecules.
(f) Gram Atom
A gram atom is a term, now less commonly used, that refers to the quantity of a chemical element whose mass in grams is numerically equal to its relative atomic mass. In modern terminology, it is simply equivalent to one mole of atoms of that element. For example, one gram atom of carbon-12 is exactly 12 grams of carbon-12.
(g) Mole
The mole is the SI unit for the amount of substance. One mole contains exactly 6.02214076 × 10²³ (Avogadro’s number) elementary entities, which may be atoms, molecules, ions, or other particles. It is a counting unit that allows chemists to work with a manageable number of particles when dealing with macroscopic quantities in the laboratory.
3.(a) What are the main applications of Avogadro’s Law?
(b) How does Avogadro’s Law explain Gay-Lussac’s Law of combining volumes?
Ans: (a) Avogadro’s Law has several key applications. It provides the foundation for determining the molar volume of gases, stating that one mole of any gas occupies the same volume at a given temperature and pressure (approximately 22.4 litres at STP). This principle is crucial in determining the molecular masses of unknown gases by comparing their densities. Furthermore, it helps in establishing the molecular formulae of gaseous compounds by relating the volumes of reacting gases to the number of molecules involved, making it indispensable for stoichiometric calculations in chemical reactions involving gases.
(b) Avogadro’s Law offers a clear particle-based explanation for Gay-Lussac’s Law of combining volumes. Gay-Lussac observed that gases react in simple whole-number volume ratios. Avogadro proposed that equal volumes of all gases, under the same conditions, contain an equal number of molecules. Therefore, the simple volume ratios observed in reactions directly correspond to simple ratios in the number of reacting molecules. For instance, the reaction where two volumes of hydrogen combine with one volume of oxygen to form two volumes of water vapour implies that two molecules of hydrogen combine with one molecule of oxygen to yield two molecules of water, perfectly aligning the macroscopic volume ratios with the microscopic molecular ratios.
4. Calculate the relative molecular masses of :
(a) Ammonium chloroplatinate
(NH4)2PtCl6
(b) Potassium chlorate,
(c)CuSO4⋅5H2O
(d)(NH4)2SO4
(e)CH3COONa
(f)CHCl3
(g)(NH4)2Cr2O7
Ans: The relative molecular mass (also known as molecular weight) is calculated by summing the atomic masses of all the atoms present in the chemical formula. The atomic masses used are from the periodic table (C=12, H=1, N=14, O=16, Na=23, S=32, Cl=35.5, K=39, Cr=52, Pt=195, Cu=63.5).
(a) Ammonium chloroplatinate, (NH₄)₂PtCl₆
Nitrogen (N): 2 atoms × 14 = 28
Hydrogen (H): 8 atoms × 1 = 8
Platinum (Pt): 1 atom × 195 = 195
Chlorine (Cl): 6 atoms × 35.5 = 213
Relative Molecular Mass = 28 + 8 + 195 + 213 = 444
(b) Potassium chlorate, KClO₃
Potassium (K): 1 atom × 39 = 39
Chlorine (Cl): 1 atom × 35.5 = 35.5
Oxygen (O): 3 atoms × 16 = 48
Relative Molecular Mass = 39 + 35.5 + 48 = 122.5
(c) Hydrated Copper Sulphate, CuSO₄·5H₂O
This formula includes 5 water molecules. We calculate the mass of CuSO₄ and the mass of 5H₂O separately and then add them.
Copper Sulphate (CuSO₄):
Copper (Cu): 1 atom × 63.5 = 63.5
Sulphur (S): 1 atom × 32 = 32
Oxygen (O): 4 atoms × 16 = 64
Mass of CuSO₄ = 63.5 + 32 + 64 = 159.5
Water of Crystallisation (5H₂O):
Hydrogen (H): 10 atoms × 1 = 10
Oxygen (O): 5 atoms × 16 = 80
Mass of 5H₂O = 10 + 80 = 90
Relative Molecular Mass = Mass of CuSO₄ + Mass of 5H₂O = 159.5 + 90 = 249.5
(d) Ammonium Sulphate, (NH₄)₂SO₄
Nitrogen (N): 2 atoms × 14 = 28
Hydrogen (H): 8 atoms × 1 = 8
Sulphur (S): 1 atom × 32 = 32
Oxygen (O): 4 atoms × 16 = 64
Relative Molecular Mass = 28 + 8 + 32 + 64 = 132
(e) Sodium Acetate, CH₃COONa
Carbon (C): 2 atoms × 12 = 24
Hydrogen (H): 3 atoms × 1 = 3
Oxygen (O): 2 atoms × 16 = 32
Sodium (Na): 1 atom × 23 = 23
Relative Molecular Mass = 24 + 3 + 32 + 23 = 82
(f) Chloroform, CHCl₃
Carbon (C): 1 atom × 12 = 12
Hydrogen (H): 1 atom × 1 = 1
Chlorine (Cl): 3 atoms × 35.5 = 106.5
Relative Molecular Mass = 12 + 1 + 106.5 = 119.5
(g) Ammonium Dichromate, (NH₄)₂Cr₂O₇
Nitrogen (N): 2 atoms × 14 = 28
Hydrogen (H): 8 atoms × 1 = 8
Chromium (Cr): 2 atoms × 52 = 104
Oxygen (O): 7 atoms × 16 = 112
Relative Molecular Mass = 28 + 8 + 104 + 112 = 252
Summary of Answers:
(a) (NH₄)₂PtCl₆ : 444
(b) KClO₃ : 122.5
(c) CuSO₄·5H₂O : 249.5
(d) (NH₄)₂SO₄ : 132
(e) CH₃COONa : 82
(f) CHCl₃ : 119.5
(g) (NH₄)₂Cr₂O₇ : 252
5. Find the :
(a) number of molecules in 73 g of HCl,
(b) weight of 0.5 mole of O2,
(c) number of molecules in 1.8 g of H2O,
(d) number of moles in 10 g of CaCO3,
(e) weight of 0.2 mole of H2gas,
(f) number of molecules in 3.2 g of SO2.
Ans: (a) 73 g of HCl
Molar mass of HCl = 36.5 g/mol
Moles = 73 ÷ 36.5 = 2 moles
Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules
(b) 0.5 mole of O₂
Molar mass of O₂ = 32 g/mol
Mass = 0.5 × 32 = 16 g
(c) 1.8 g of H₂O
Molar mass of H₂O = 18 g/mol
Moles = 1.8 ÷ 18 = 0.1 mole
Molecules = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules
(d) 10 g of CaCO₃
Molar mass of CaCO₃ = 100 g/mol
Moles = 10 ÷ 100 = 0.1 mole
(e) 0.2 mole of H₂ gas
Molar mass of H₂ = 2 g/mol
Mass = 0.2 × 2 = 0.4 g
(f) 3.2 g of SO₂
Molar mass of SO₂ = 64 g/mol
Moles = 3.2 ÷ 64 = 0.05 mole
Molecules = 0.05 × 6.022 × 10²³ = 3.011 × 10²² molecules
6. Which of the following would weigh most :
(a) 1 mole of H2O,
(b) 1 mole of CO2,
(c) 1 mole of NH3,
(d) 1 mole of CO ?
Ans: (b) 1 mole of CO2,
7. Which of the following contains maximum number of molecules :
(a) 4 g of O2,
(b) 4 g of NH3,
(c) 4 g of CO2,
(d) 4 g of SO2?
Ans: Step 1: Understanding the Concept
The number of molecules in a given mass of a substance is directly proportional to the number of moles it contains. This is because one mole of any substance contains the same number of molecules, which is Avogadro’s number (6.022×1023 ).
Therefore, to find which sample has the most molecules, we simply need to find which one has the highest number of moles.
The number of moles (n) is calculated as:
n= Given Mass (g) / Molar Mass (g/mol)
Since the given mass is the same (4 g) for all the substances, the substance with the smallest molar mass will have the largest number of moles, and hence, the maximum number of molecules.
Step 2: Calculate the Molar Mass of Each Substance
(a) Oxygen gas (O₂):
Molar mass = 2×16=32 g/mol
(b) Ammonia (NH₃):
Molar mass =
14+(3×1)=17 g/mol
(c) Carbon dioxide (CO₂):
Molar mass =
12+(2×16)=44 g/mol
(d) Sulfur dioxide (SO₂):
Molar mass =
32+(2×16)=64 g/mol
Step 3: Calculate the Number of Moles in 4 g of Each Substance
(a) Moles of O₂:
n= 432= 0.125 mol
(b) Moles of NH₃:
n= 174 ≈0.235 mol
(c) Moles of CO₂:
n= 444 ≈0.091 mol
(d) Moles of SO₂:
n= 64 =0.0625 mol
Step 4: Compare the Number of Moles
Comparing the calculated moles:
NH3(0.235 mol)>O2(0.125 mol)>CO2(0.091 mol)>SO2(0.0625 mol)
Ammonia (NH₃) has the highest number of moles.
Step 5: Final Conclusion
Since 4 g of NH₃ has the largest number of moles, it contains the maximum number of molecules.
(b) 4 g of NH3
8. Calculate the number of :
(a) particles in 0.1 mole of any substance,
(b) hydrogen atoms in 0.1 mole of H2SO4,
(c) molecules in one kg of calcium chloride.
Ans: (a) Particles in 0.1 mole of any substance
In chemistry, the mole is a fundamental unit for counting particles. One mole of any substance, whether it’s atoms, molecules, or ions, always contains the same number of particles, known as Avogadro’s constant. This value is approximately 6.022 × 10²³.Therefore, for 0.1 mole of a substance, the number of particles is calculated by multiplying Avogadro’s constant by 0.1.
Number of particles = 0.1 × (6.022 × 10²³) = 6.022 × 10²²
Conclusion: There are 6.022 × 10²² particles in 0.1 mole of any substance.
(b) Hydrogen atoms in 0.1 mole of H₂SO₄
To find the number of hydrogen atoms, we first need to find the number of sulfuric acid (H₂SO₄) molecules. As established, 0.1 mole of any substance contains 6.022 × 10²² fundamental units. In this case, the units are H₂SO₄ molecules.
Number of H₂SO₄ molecules = 0.1 × (6.022 × 10²³) = 6.022 × 10²² molecules
Now, looking at the chemical formula H₂SO₄, we see that each molecule contains 2 hydrogen (H) atoms. To find the total number of hydrogen atoms, we multiply the total number of molecules by 2.
Total H atoms = (6.022 × 10²² molecules) × 2 = 1.2044 × 10²³ atoms
Conclusion: There are approximately 1.204 × 10²³ hydrogen atoms in 0.1 mole of H₂SO₄.
(c) Molecules in one kg of calcium chloride
First, we determine the molar mass of calcium chloride (CaCl₂).
Atomic mass of Calcium (Ca) = 40 g/mol
Atomic mass of Chlorine (Cl) = 35.5 g/mol
Molar mass of CaCl₂ = 40 + 2(35.5) = 40 + 71 = 111 g/mol.
This means that one mole of CaCl₂ has a mass of 111 grams.
The given mass is 1 kilogram, which is 1000 grams. To find out how many moles this represents, we use the formula:
Number of moles = Given mass / Molar mass
Number of moles = 1000 g / 111 g/mol ≈ 9.009 moles
Finally, we calculate the number of molecules by multiplying the number of moles by Avogadro’s constant.
Number of molecules = 9.009 moles × (6.022 × 10²³ molecules/mole)
Number of molecules ≈ 5.426 × 10²⁴ molecules
Conclusion: There are approximately 5.426 × 10²⁴ molecules in one kilogram of calcium chloride.
9. How many grams of :
(a) Al are present in 0.2 mole of it,
(b) HCl are present in 0.1 mole of it ?
(c) H2O are present in 0.2 mole of it,
(d) CO2is present in 0.1 mole of it ?
Ans: The mass of a substance is calculated using the formula:
Mass = Number of moles × Molar mass
(a) Mass of Aluminium (Al)
Number of moles = 0.2 mol
Molar mass of Al = 27 g/mol
Mass = 0.2 mol × 27 g/mol = 5.4 grams
(b) Mass of Hydrochloric Acid (HCl)
Number of moles = 0.1 mol
Molar mass of HCl = Atomic mass of H + Atomic mass of Cl = 1 + 35.5 = 36.5 g/mol
Mass = 0.1 mol × 36.5 g/mol = 3.65 grams
(c) Mass of Water (H₂O)
Number of moles = 0.2 mol
Molar mass of H₂O = (2 × Atomic mass of H) + Atomic mass of O = (2 × 1) + 16 = 18 g/mol
Mass = 0.2 mol × 18 g/mol = 3.6 grams
(d) Mass of Carbon Dioxide (CO₂)
Number of moles = 0.1 mol
Molar mass of CO₂ = Atomic mass of C + (2 × Atomic mass of O) = 12 + (2 × 16) = 44 g/mol
Mass = 0.1 mol × 44 g/mol = 4.4 grams
10.(a) The mass of 5.6 litres of a certain gas at STP is 12 g. What is the relative molecular mass or molar mass of the gas ?
(b) Calculate the volume occupied at S.T.P. by 2 moles of SO2.
Ans: (a)Volume at STP = 5.6 L
Mass = 12 g
Moles of gas =
5.6/22.4=0.25mol
Molar mass =
12/0.25=48g/mol
Relative molecular mass = 48
(b)1 mole of gas at STP occupies 22.4 L
2 moles of SO2
occupy =2×22.4=44.8 L
Volume = 44.8 L
11. Calculate the number of moles of :
(a) CO2which contain 8.00 g of O2,
(b) methane in 0.80 g of methane.
Ans: (a) Moles of CO₂ containing 8.00 g of O₂
Molar mass of O₂ = 32 g/mol
Moles of O₂ = 8.00 / 32 = 0.25 mol
Each mole of O₂ has 2 moles of O atoms → 0.25 × 2 = 0.5 mol O atoms
Each CO₂ has 2 O atoms → Moles of CO₂ = 0.5 / 2 = 0.25 mol
(b) Moles of methane in 0.80 g
Molar mass of CH₄ = 16 g/mol
Moles = 0.80 / 16 = 0.05 mol
12. Calculate the actual mass of :
(a) an atom of oxygen,
(b) an atom of hydrogen,
(c) a molecule of NH3,
(d) the atom of silver,
(e) the molecule of oxygen,
(f) 0.25 gram atom of calcium.
Ans: (a) Actual mass of an atom of oxygen
= Atomic mass / Avogadro’s number
= 16 g/mol / (6.022 × 10²³ atoms/mol)
≈ 2.656 × 10⁻²³ g
(b) Actual mass of an atom of hydrogen
= 1 g/mol / (6.022 × 10²³ atoms/mol)
≈ 1.660 × 10⁻²⁴ g
(c) Actual mass of a molecule of NH₃
Molecular mass of NH₃ = 14 + 3(1) = 17 g/mol
= 17 g/mol / (6.022 × 10²³ molecules/mol)
≈ 2.823 × 10⁻²³ g
(d) Actual mass of an atom of silver
= 108 g/mol / (6.022 × 10²³ atoms/mol)
≈ 1.793 × 10⁻²² g
(e) Actual mass of a molecule of oxygen (O₂)
Molecular mass of O₂ = 32 g/mol
= 32 g/mol / (6.022 × 10²³ molecules/mol)
≈ 5.313 × 10⁻²³ g
(f) Mass of 0.25 gram atom of calcium
Gram atomic mass of Calcium (Ca) = 40 g
Mass = 0.25 × 40 g
= 10 g
13. Calculate the mass of 0.1 mole of each of the following:
(a) CaCO3,
(b) Na2SO4· 10H2O,
(c) CaCl2,
(d) Mg.
(Ca = 40, Na = 23, Mg = 24, S = 32, C = 12, Cl = 35.5, O = 16, H = 1)
Ans: (a) 10 g (b) 32.2 g (c) 11.1 g (d) 2.4 g
14. Calculate the number of oxygen atoms in 0.10 mole of Na2CO3· 10H2O.
Ans: Step 1: Determine number of oxygen atoms in one formula unit
Formula: Na₂CO₃·10H₂O
Na₂CO₃ contains 3 O atoms
10H₂O contains 10 × 1 = 10 O atoms? Wait — each H₂O has 1 O atom, so 10 H₂O has 10 O atoms.
Total O atoms per formula unit = 3 (from CO₃) + 10 (from 10H₂O) = 13 O atoms.
Step 2: Calculate total O atoms in 0.10 mol
Number of formula units in 0.10 mol
= 0.10 × (6.022 × 10²³)
= 6.022 × 10²² formula units.
Each formula unit has 13 O atoms, so:
O atoms = 6.022 × 10²² × 13
= 78.286 × 10²²
= 7.8 × 10²³ O atoms.
Final Answer:7.8×10 *23
15. What mass of Ca will contain the same number of atoms as are present in 3.2 g of S ?
Ans: Step 1: Find moles of sulfur (S).
Molar mass of S = 32 g/mol
Given mass = 3.2 g
Moles of S = 3.2/32=0.1mol
Step 2: Number of atoms same → moles same.
So, moles of Ca needed = 0.1 mol
Step 3: Mass of Ca
Molar mass of Ca = 40 g/mol
Mass = 0.1×40=4.0 g
Final answer: 4.0 g
16. Calculate the number of atoms in each of the following :
(a) 52 moles of He,
(b) 52 amu of He,
(c) 52 g of He.
Ans: Step 1 – Recall constants
1 mole of atoms = 6.022×1023 atoms
Molar mass of He = 4 g/mol
Mass of 1 He atom ≈ 4 amu
(a) 52 moles of He
Number of atoms=52×6.022×1023=3.131×1025 atoms
(b) 52 amu of He
Number of atoms= 4 amu/atom/52 amu =13 atoms
(c) 52 g of He
Moles=52/4=13 moles
Atoms=13×6.022×1023=7.829×10/24 atoms
Final answers:(a) 3.131×10/25 atoms
(b) 13 atoms
(c) 7.829×10 24 atoms
17. Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.
Ans: Step 1: Write the chemical formula of sodium carbonate.
The formula is Na₂CO₃.
Step 2: Calculate the molecular mass of Na₂CO₃.
Mass of 2 Sodium (Na) atoms = 2 × 23 = 46 u
Mass of 1 Carbon (C) atom = 1 × 12 = 12 u
Mass of 3 Oxygen (O) atoms = 3 × 16 = 48 u
Molecular mass of Na₂CO₃ = 46 + 12 + 48 = 106 u
So, the gram molecular mass of Na₂CO₃ is 106 grams/mol. This means 106 grams of sodium carbonate contain 1 mole of Na₂CO₃ molecules.
Step 3: Find the number of moles in 5.3 grams of Na₂CO₃.
Number of moles (n) = (Given Mass) / (Molar Mass)
n = 5.3 g / 106 g/mol
n = 0.05 moles
Step 4: Calculate the number of molecules of Na₂CO₃.
1 mole of any substance contains 6.022 × 10²³ molecules (Avogadro’s number).
Number of molecules = Number of moles × Avogadro’s number
= 0.05 mol × 6.022 × 10²³ molecules/mol
= 3.011 × 10²² molecules
Step 5: Determine the number of atoms of each kind.
Now, we analyze one molecule of Na₂CO₃:
One molecule of Na₂CO₃ contains:
2 atoms of Sodium (Na)
1 atom of Carbon (C)
3 atoms of Oxygen (O)
Therefore, the total number of atoms of each element is:
Number of Sodium (Na) atoms = (Number of molecules) × 2
= (3.011 × 10²²) × 2
= 6.022 × 10²² atoms
Number of Carbon (C) atoms = (Number of molecules) × 1
= (3.011 × 10²²) × 1
= 3.011 × 10²² atoms
Number of Oxygen (O) atoms = (Number of molecules) × 3
= (3.011 × 10²²) × 3
= 9.033 × 10²² atoms
18.(a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH2)
2].[O = 16; N = 14; C = 12; H = 1]
(b) Calculate the volume occupied by 320 g of sulphur dioxide at STP [S = 32; O = 16]
Ans: (a) Mass of nitrogen in 5 kg of urea
Step 1: Molar mass of Urea, CO(NH₂)₂
C = 12 g/mol
O = 16 g/mol
N = 14 × 2 = 28 g/mol
H = 1 × 4 = 4 g/mol
Molar mass = 12 + 16 + 28 + 4 = 60 g/mol
Step 2: Mass of N in 1 mol urea
1 mol urea (60 g) contains 28 g of N.
Step 3: Mass of N in 5 kg (5000 g) urea
Mass of N = (28 / 60) × 5000 g
= (28 × 5000) / 60
= 140000 / 60
= 2333.33 g or 2.33 kg
Answer (a): 2.33 kg
(b) Volume of 320 g of SO₂ at STP
Step 1: Molar mass of SO₂
S = 32 g/mol
O = 16 × 2 = 32 g/mol
Molar mass = 32 + 32 = 64 g/mol
Step 2: Moles in 320 g of SO₂
Moles = Given mass / Molar mass
= 320 g / 64 g mol⁻¹
= 5 moles
Step 3: Volume at STP
1 mole at STP = 22.4 L
5 moles = 5 × 22.4 L
= 112 L
Answer (b): 112 L
19.(a) What do you understand by the statement that ‘vapour density of carbon dioxide is 22’ ?
(b) Atomic mass of chlorine is 35.5. What is its vapour density ?
Ans:(a)Vapour density of carbon dioxide is 22 means that the mass of a certain volume of carbon dioxide gas is 22 times the mass of an equal volume of hydrogen gas measured under the same conditions of temperature and pressure.
(b)Vapour density (VD) of a gas = Molecular mass/2
For chlorine (Cl₂), molecular mass = 2×35.5=71
So, VD = 71/2=35.52
Ans:
(a) Carbon dioxide is 22 times heavier than hydrogen.
(b) Vapour density of chlorine is 35.5.
20. What is the mass of 56 cm3of carbon monoxide at STP?
[C = 12; O = 16]
Ans: Step 1: Moles of gas
Volume = 56 cm³
Molar volume = 22400 cm³/mol
Moles = 56/22400=0.0025
Step 2: Molar mass of CO
Molar mass = 12 + 16 = 28 g/mol
Step 3: Mass
Mass = 0.0025×28=0.07g
Final Answer: 0.07 g
21. Determine the no. of molecules in a drop of water which weighs 0.9 g.
Ans: Step 1: Understand what we are given and what we need.
We know:
Mass of the water drop = 0.9 grams
We need to find the number of molecules.
We also need to recall a fundamental concept from chemistry:
The molar mass of water (H₂O) is 18 g/mol. This means that 18 grams of water contain one mole of water molecules.
Avogadro’s number (Nₐ) is 6.022 × 10²³.
So, our plan is to find out how many moles are in 0.9 grams of water, and then multiply that by Avogadro’s number to find the number of molecules.
Step 2: Calculate the number of moles.
The formula for the number of moles (n) is:
n=Given Mass/Molar Mass
Plugging in our values:
N =0.9g/18g/mol=0.05moles
So, the drop of water contains 0.05 moles of H₂O molecules.
Step 3: Calculate the number of molecules.
Now, we use Avogadro’s number to convert moles to molecules.
Number of Molecules=0.05mol×(6.022×1023molecules/mol)
Number of Molecules=0.05mol×(6.022×1023 molecules/mol)
Let’s do the multiplication:
0.05×6.022=0.3011
So,Number of Molecules=0.3011×10/23
To write this in proper scientific notation, we can express it as:
Number of Molecules=3.011×1022
Final Answer:
3.011×1022
This is the number of individual water molecules in a 0.9-gram drop of water. It’s a staggeringly large number, which shows just how tiny a single molecule is
22. The molecular formula for elemental sulphur is S8. In a sample of 5.12 g of sulphur :
(a) How many moles of sulphur are present,
(b) How many molecules and atoms are present ?
Ans: Step 1: Calculate the molar mass of S₈
Atomic mass of S = 32 g/mol
Molar mass of S₈ = 8 × 32 = 256 g/mol
(a) Moles of sulphur
Moles=Mass5.12/256=0.02 mol
Moles= 256/5.12 =0.02 mol
(b) Number of molecules of S₈
Molecules=Moles×N /A
=0.02×6.022×10 23=1.2044×1022 molecules
Number of sulphur atoms
Each S₈ molecule has 8 atoms, so:
Atoms=1.2044×10 22 ×8=9.635×10 *22 atoms
23. If phosphorus is considered to contain P4 molecules, then calculate the number of moles in 100 g of phosphorus?
Ans: Step 1: Understand the given information
We are told that phosphorus exists as P₄ molecules. This means the basic unit for our calculation is a molecule made of 4 phosphorus atoms.
The mass of the phosphorus sample given is 100 g.
Step 2: Determine the molar mass of P₄
The atomic mass of phosphorus (P) is 31 g/mol.
Since one P₄ molecule contains 4 phosphorus atoms, its molar mass is:
Molar mass of P₄ = 4 × 31 g/mol = 124 g/mol.
This means that one mole of P₄ molecules has a mass of 124 grams.
Step 3: Apply the mole formula
Number of moles=Given mass/Molar mass
Number of moles= Molar mass/Given mass
Step 4: Substitute the values and calculate
Number of moles of P₄=100g/124 g/mol
Number of moles of P₄= 0.80645mol
Number of moles of P₄=0.80645mol
Step 5: State the final answer
Rounding off to an appropriate number of significant figures, the number of moles in 100 g of phosphorus (P₄) is:0.806
24. Calculate :
(a) the gram molecular mass of chlorine if 308 cm3 of it at STP weighs 0.979 g,
(b) the volume of 4 g of H2 at 4 atmosphere,
(c) the mass of oxygen in 2.2 litres of CO2 at STP.
Ans: 24. (a) Gram Molecular Mass of Chlorine
Given:Volume of chlorine gas at STP = 308 cm³
Mass of this volume = 0.979 g
Step 1: Recall the principle.
The gram molecular mass of any gas occupies 22.4 litres at STP.
Step 2: Convert the given volume to litres.
We know that 1000 cm³ = 1 litre.
So, 308 cm³ = 308 / 1000 = 0.308 litres.
Step 3: Find the mass of 22.4 litres.
If 0.308 litres of chlorine weighs 0.979 g,
Then, 1 litre of chlorine would weigh 0.979 g.
Therefore, 22.4 litres (the molar volume) would weigh
0.979/0.308×22.4g.
Step 4: Perform the calculation.
First,
0.979/0.308=3.17857
Then,
3.17857×22.4=71.2 g (approximately)
Final Answer for (a):
The gram molecular mass of chlorine is 71.2 g.
(b) Volume of 4 g of H₂ at 4 atmosphere
Given:
Mass of Hydrogen gas (H₂) = 4 g
Pressure = 4 atm
Temperature is assumed to be constant (not specified, so we assume it’s the same as STP for the application of Boyle’s Law).
Step 1: Find the number of moles of H₂.
Molecular mass of H₂ = 2 g/mol.
Number of moles (n) =
Given Mass
Molar Mass=42=2
Molar Mass
Given Mass = 24 =2 moles.
Step 2: Find the volume at STP (1 atm).
At STP, 1 mole of any gas occupies 22.4 litres.
So, volume of 2 moles at STP (V₁) = 2×22.4=44.8 litres.
Pressure at STP (P₁) = 1 atm.
Step 3: Find the volume at the new pressure (4 atm) using Boyle’s Law.
Boyle’s Law states: P₁V₁ = P₂V₂ (at constant temperature).
We have:
P₁ = 1 atm
V₁ = 44.8 litres
P₂ = 4 atm
V₂ = ? (This is what we need to find)
Step 4: Apply Boyle’s Law.
P1V1=P2V2
1×44.8=4×V 2
V2=44.8/4 =11.2 litres.
Final Answer for (b):
The volume of 4 g of H₂ at 4 atmosphere is 11.2 litres.
(c) Mass of Oxygen in 2.2 litres of CO₂ at STP
Given:Volume of CO₂ at STP = 2.2 litres
Step 1: Find the number of moles of CO₂.
At STP, 22.4 litres of any gas = 1 mole.
So, number of moles in 2.2 litres = 22.4=22/
224=11/112 moles.
Step 2: Analyze the composition of CO₂.
One molecule of CO₂ contains one atom of Carbon and two atoms of Oxygen.
Therefore, 1 mole of CO₂ contains 2 moles of Oxygen atoms.
Step 3: Find the number of moles of Oxygen.
Moles of O = 2×
moles of CO₂=2×moles of CO₂=2× 112/11 = 112/22= 56/11moles.
Step 4: Calculate the mass of Oxygen.
Atomic mass of Oxygen (O) = 16 g/mol.
Since oxygen exists as O₂ in this context, but the mass calculation remains the same for the atoms, the mass of oxygen is:
Mass = Number of moles of O atoms × Molar mass of O
Mass = 11/56×16 g
Step 5: Perform the calculation.
11/56×16=11×16/56=176/56=3.142=3.142 g (approximately)
Final Answer for (c):
The mass of oxygen in 2.2 litres of CO₂ at STP is 3.14 g.
25. A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10−12g, calculate the number of carbon atoms in the signature.
Ans: Calculating the Number of Atoms in a Microscopic Carbon Sample
Let’s determine how many atoms are in an incredibly small speck of carbon, weighing just one picogram (which is 0.000000000001 grams or 10⁻¹² g).
Step 1: Find the Number of Moles of Carbon
First, we need to understand that a “mole” is a chemist’s counting unit, similar to how a “dozen” means 12 of something. One mole of any substance contains exactly Avogadro’s number of particles (atoms, molecules, etc.), which is approximately 6.022 × 10²³.
The number of moles (*n*) is calculated by dividing the given mass of a substance by its molar mass (the mass of one mole of that substance).
Given Mass of Carbon: 10⁻¹² grams
Molar Mass of Carbon: 12 grams per mole (from the periodic table)
The calculation is:
*n* = (10⁻¹² g) / (12 g/mol) = 8.33 × 10⁻¹⁴ moles
So, this tiny carbon sample amounts to 0.0000000000000833 moles.
Step 2: Find the Number of Atoms
Now that we know how many moles we have, we can find the actual number of atoms. We simply multiply the number of moles by Avogadro’s number, which tells us how many atoms are in each mole.
Number of Moles (*n*): 8.33 × 10⁻¹⁴ mol
Avogadro’s Number (Nₐ): 6.022 × 10²³ atoms/mol
The calculation is:
N = (8.33 × 10⁻¹⁴ mol) × (6.022 × 10²³ atoms/mol)
Let’s break that multiplication down:
Multiply the numbers: 8.33 × 6.022 ≈ 50.15
Add the exponents: 10⁻¹⁴ × 10²³ = 10⁽⁻¹⁴ ⁺ ²³⁾ = 10⁹
This gives us a preliminary result of 50.15 × 10⁹
To express this properly in scientific notation, we adjust it to 5.015 × 10¹⁰.
Final Answer: Therefore, a picogram (10⁻¹² g) sample of carbon contains approximately 5.02 × 10¹⁰ atoms.
To put this enormous number into perspective, even in such an unimaginably small sample, there are still over 50 billion individual carbon atoms.
26. An unknown gas shows a density of 3 g per litre at 273°C and 1140 mm Hg pressure. What is the gram molecular mass of this gas ?
Ans: Step 1: Given data
Density d=3 g/L
Temperature
T=273
C=273+273=546 K
Pressure
P=1140 mm Hg
Gas constant
Step 2: Convert pressure to atm
P=1140/760 atm=1.5 atm
Step 3: Use the relationship
For an ideal gas:
PM=dRT
where
M = molar mass (g/mol).
M=dRTP
Step 4: Substitute values
M=3×0.0821×546/1.5
M= 1.5
M=134.4798/1.5≈89.65 g/mol
Final Answer:89.65
27. Cost of Sugar (C12H22O11) is ₹ 40 per kg; calculate its cost per mole.
Ans: Step 1: Find the molar mass of sugar (C₁₂H₂₂O₁₁)
Carbon (C): 12×12=144 g/mol
Hydrogen (H): 22×1=22 g/mol
Oxygen (O): 11×16=176 g/mol
Molar mass = 144+22+176=342 g/mol
Step 2: Convert price per kg to price per mole
1 kg = 1000 g
Cost per gram = 1000/40 =₹0.04 per gram
Cost of 342 g (1 mole) = 342×0.04=₹13.68
Final Answer:13.68
28. Which of the following weighs the least ?
(a) 2 g atom of N
(b) 3 ×10251 atoms of carbon
(c) 1 mole of sulphur
(d) 7 g silver
Ans: (d) 7 g silver
29. Four grams of caustic soda contains :
(a) 6.02 ×1023 atoms of it,
(b) 4 g atom of sodium,
(c)6.02×1022 molecules
(d) 4 moles of NaOH.
Ans: The correct answer is (c) 6.02 × 10²² molecules.
4 grams of NaOH corresponds to 0.1 moles.
Multiplying 0.1 moles by Avogadro’s number (6.02 × 10²³ molecules/mol) gives 6.02 × 10²² molecules.
The other options are incorrect due to miscalculations in the number of atoms, moles, or gram-atoms.
30. The number of molecules in 425 g of ammonia is :
(a)1.0×1023
(b)1.5×1023
(c)2.0×1023
(d)3.5×1023
Ans: (b)1.5×1023
31. Correct the statements, if required.
(a) One mole of chlorine contains 6.023×1022atoms of chlorine,
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with two volumes of oxygen will give two volumes of water vapour,
(c) Relative atomic mass of an element is the number of times one molecule of an element is heavier than 1/12 the mass of an atom of carbon [C12],
(d) Under the same conditions of the temperature and pressure, equal volumes of all gases contain the same number of atoms.
Ans:
(a) Incorrect.
Corrected statement: One mole of chlorine gas (Cl₂) contains
6.023×1023
molecules, and each molecule has 2 atoms, so total atoms = 1.2046×1024
(b) Incorrect.
Corrected statement: Under similar conditions, two volumes of hydrogen combine with one volume of oxygen to give two volumes of water vapour.
(c) Incorrect.
Corrected statement: Relative atomic mass of an element is the number of times one atom of the element is heavier than
1/12
the mass of a carbon-12 atom.
(d) Incorrect.
Corrected statement: Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (not atoms).
