Mole Concept & Stoichiometry Miscellaneous Exercises
Solutions for Chemistry, Class 10, ICSE
Miscellaneous Exercises Long Answer Type
4 questionsAnswer:
Information conveyed by formula [H2O] —
- One molecule of water (H2O) is made of two atoms of Hydrogen and one atom of Oxygen.
- As atomic weight of hydrogen is 1 and that of oxygen is 16. Therefore, ratio by weight of hydrogen and oxygen is = =
- Molecular weight of H2O is 2H + O = 2 + 16 = 18g.
Answer:
(a) Stoichiometry measures quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry.
(b) Atomicity is the number of atoms in a molecule of an element.
Atomicity of hydrogen is 2, phosphorus is 4 and sulphur is 8.
(c) Difference between N2 and 2N
N2 | 2N |
---|---|
It means 1 molecule of nitrogen. | It means two atoms of nitrogen. |
It can exist independently. | It cannot exist independently. |
Answer:
(a) The applications of Avogadro's Law are:
- It explains Gay-Lussac's law.
- It predicts atomicity of gases.
- It determines the molecular formula of gases.
- It determines the relation between molecular mass and vapour density.
- It gives the relationship between gram molecular mass and gram molar volume.
(b) According to Avogadro's law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another. This is what Gay Lussac's Law says.
Answer:
(a) The relative atomic masses of any element is the weighted average of the relative atomic masses of it's natural isotopes. Chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio 3 : 1.
The average relative atomic mass of Cl = = 35.5
(b) 6.022 × 1023
(c) The molar volume of a gas is 22.4 dm3 (litre) or 22400 cm3 (ml) at S.T.P.
Miscellaneous Exercises Multiple Choice Type
17 questionsAnswer:
7 g of silver
Reason —
- Weight of 1 g atom of N = 14 g
∴ weight of 2 g atom of N = 28 g
- 6.022 × 1023 atoms of C weigh = 12 g
∴ 3 × 1025 atoms will weigh = × 3 × 1025 = 597.7 g
1 mole of sulphur weighs = 32 g
7 g of silver
The weight computed in all other options is greater than the weight in option (d). Hence, 7 grams of silver weighs the least.
Answer:
6.02 × 1022 molecule
Reason
Molar mass of NaOH = Na + O + H = 23 + 16 + 1 = 40 g
40 g of NaOH contains 6.022 × 1023 molecules
∴ 4 g of NaOH contains
= x 4
= 6.02 × 1022 molecules
Answer:
1.5 × 1023
Reason
Molar mass of ammonia = N + 3H = 14 + 3 = 17 g
The number of molecules in 17 g of ammonia = 6.022 × 1023
∴ No. of molecules in 4.25 g of ammonia
= x 4.25
= 1.5 × 1023
A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:
5
10
15
20
Answer:
10
Reason
Vapour density of gas =
Molecular weight = 2 x Vapour density
= 2 x 5 = 10 g
Hence, relative molecular mass of gas is 10 g
Answer:
molecular mass
Reason
Relative V.D. =
According to Avogadro's Law, volumes at the same temperature and pressure may be substituted by molecules. Hence,
Relative V.D. =
(Molecule of hydrogen contains 2 atoms) By multiplying both sides by 2
2 x Relative V.D = Relative molecular mass of a gas or vapour
So, twice the vapour density gives relative molecular mass.
Answer:
C3H6O3
Reason — The empirical formula of a compound is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The given empirical formula is CH2O
The simplest ratio of given compound, is 1:2:1
So, C3H6O3 has simplest ratio in multiples of 3
i.e., 3 x CH2O = C3H6O3
Whereas C2H4O, C4H3O2 and C4H6O2 has simplest ratio 2:4:1, 4:3:2 and 2:3:1 respectively. This cannot be derived from the simplest ratio 1:2:1 by multiplying it with any whole number value of n.
Answer:
29
Reason
Given relative molecular mass of (C4H10) is 58
2 x Relative V.D. =
2 x Relative V.D = Relative molecular mass of a gas or vapour.
Rearranging the above formula
Relative V.D =
Relative V.D = = 29
Hence the vapour density of (C4H10) is 29
Answer:
P3Q6
Reason
Molecular formula = empirical formula x n
Hence the molecular formula is,
= 3 x empirical formula
= 3 x PQ2 = P3Q6
Answer:
1:1
Reason
Hydrogen (H2):
Molecular mass of H2 = 2 g/mol
Moles of H2 = = 1 mole
Molecules = 1 × Avogadro’s number = 6.022 × 1023
Oxygen (O2):
Molecular mass of O2 = 32 g/mol
Moles of O2 = = 1 mole
Molecules = 1 × Avogadro’s number = 6.022 × 1023
Hence the ratio is 1:1
Answer:
Both P and R
Reason —
One mole of sulphur dioxide (SO2) contains 6.02 x 1023 molecules and is equivalent to its gram molecular mass.
At Standard Temperature and Pressure (STP),
1 mole of any ideal gas = 22.4 litres
So, 1 mole of sulphur dioxide (SO2) gas occupies 22.4 litres at STP.
Assertion (A): According to gas equation,
=
Reason (R): On combining Boyle's law and Charle's law, 'volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature."
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
Both A and R are true and R is the correct explanation of A.
Explanation — On combining Boyle's law and Charle's law we conclude that 'volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature'.
V= × constant
or = K (constant)
If the volume of a given mass of a gas changes from V1 to V2 pressure from P1 to P2 and temperature from T1 to T2 then,
= =K (constant) ⟶ Gas Equation
Hence Both assertion(A) and reason(R) are true and R is the correct explanation of A.
Assertion (A): The absolute scale of temperature is the Kelvin scale.
Reason (R): Kelvin scale starts from 0°C.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
A is true but R is false.
Explanation — A temperature scale with absolute zero (zero kelvin) as the starting point is called the absolute scale or the kelvin scale. Hence the assertion (A) is true.
Kelvin scale starts from zero kelvin or -273°C. Hence reason (R) is false.
Assertion (A): Triatomic molecules consist of four atoms.
Reason (R): Monoatomic molecules contain only one atom.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
A is false but R is true.
Explanation — Triatomic molecule is composed of three similar atoms.
Example: Ozone gas (O3). Hence the assertion (A) is false.
Monoatomic molecule is composed of only one atom.
Examples: Inert gases like Helium, Neon, Argon, etc.
Hence the reason (R) is true.
Assertion (A): One litre of hydrogen weighs the same as one litre of oxygen.
Reason (R): One litre of hydrogen contains the same number of molecules as one litre of oxygen.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
A is false but R is true.
Explanation — At standard temperature and pressure (STP), one litre of any ideal gas contains the same number of molecules (Avogadro’s law). But the masses differ because the molar masses of the gases are different. Hence the assertion (A) is false.
Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules". This means that one litre of hydrogen contains the same number of molecules as are present in one litre of oxygen or any other gas. Hence the reason (R) is true
Assertion (A): Atomic mass is expressed in atomic mass unit.
Reason (R): Atomic mass is defined as times the mass of carbon-12.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
Both A and R are true but R is not the correct explanation of A.
Explanation — Atomic mass is expressed in atomic mass units [a.m.u.].
The relative atomic mass or atomic weight of an element is the number of times one atom of the element is heavier than times of the mass of an atom of carbon-12.
Relative atomic mass =
Therefore, both assertion (A) and reason (R) are true.
Atomic mass is expressed in atomic mass unit because atoms being extremely small cannot be weighed directly and conventional units of mass like kilograms or grams are not suitable for measuring them.
Hence, both assertion (A) and reason (R) are true but reason (R) does not correctly explain assertion (A).
Assertion (A): Sulphur is an octatomic molecule.
Reason (R): The number of atoms in a molecule of an element is known as atomicity.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
Both A and R are true but R is not the correct explanation of A.
Explanation — Assertion (A) is true because sulphur molecule (S8) is composed of eight similar atoms, so it is called Octatomic molecule.
Atomicity is defined as the number of atoms in a molecule of an element.
Therefore, both assertion (A) and reason (R) are true.
However, the reason (R) does not explain why Sulphur is an octatomic molecule.
Hence, both assertion (A) and reason (R) are true but reason (R) does not correctly explain assertion (A).
Assertion (A): One mole of a gas occupies 24.4 litres at S.T.P.
Reason (R): The mass of one mole of a gas is equal to its molecular mass.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Answer:
A is false but R is true.
Explanation — At S.T.P. (Standard Temperature and Pressure), one mole of a gas actually occupies 22.4 litres, not 24.4 litres, so Assertion (A) is false.
Reason (R) is true because by definition, the mass of one mole of a substance equals its molecular mass in grams.
Miscellaneous Exercises Short Answer Type
4 questionsAnswer:
(a) Vapour density is defined as the ratio between the masses of equal volumes of gas (or vapour) and hydrogen under the same conditions of temperature and pressure.
(b) The molar volume of a gas is the volume occupied by one gram-molecular mass or by one mole of the gas at S.T.P. It is equal to 22.4 dm3.
(c) The Relative atomic mass of an element is the number of times one atom of the element is heavier than times of the mass of an atom of carbon-12.
(d) The Relative molecular mass of an element or a compound is the number that represents how many times one molecule of the substance is heavier than of the mass of an atom of carbon-12.
(e) Avogadro's number is defined as the number of atoms present in 12 g (gram atomic mass) of C-12 isotope, i.e. 6.022 x1023 atoms.
(f) The quantity of the element which weighs equal to it's gram atomic mass is called one gram atom of that element
(g) A Mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.
Answer:
(a) Gay-Lussac's Law of combining volumes — When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.
(b) Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."
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.
Answer:
(a) Avogadro's Law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."
Considering equal volumes of hydrogen and helium,
volume of hydrogen gas = volume of helium gas
According to Avogadro's Law:
n molecules of hydrogen = n molecules of helium gas
i.e., nH2 = nHe
1 molecule of hydrogen has 2 atoms of hydrogen and 1 molecule of helium has 1 atom of helium
∴ 2H = He
∴ atoms in hydrogen are double the atoms of helium.
(b) Since, the volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.
(c) According to Boyle's law, the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature. When we inflate a balloon, the volume of air keeps increasing and at the same time the pressure of air also increases due to which balloon inflates. As pressure and volume of air increase simultaneously, hence this seems to violate Boyle's law.
Answer:
The empirical formula of a compound is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.
Miscellaneous Exercises Very Short Answer Type
3 questionsComplete the following blanks in the equation as indicated.
CaH2 (s) + 2H2O (aq) ⟶ Ca(OH)2 (s) + 2H2 (g)
(a) Moles: 1 mole + ............... ⟶ ............... + ...............
(b) Grams: 42g + ............... ⟶ ............... + ...............
(c) Molecules: 6.02 x 1023 + ............... ⟶ ............... + ...............
Answer:
(a) Moles: 1 mole + 2 mole ⟶ 1 mole + 2 mole
(b) Grams: 42g + 36g ⟶ 74g + 4g
(c) Molecules: 6.02 x 1023 + 12.04 × 1023 ⟶ 6.02 x 1023 + 12.04 × 1023
Correct the statements, if required
(a) One mole of chlorine contains 6.023 × 1023 atoms 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 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.
Answer:
(a) One mole of chlorine contains 6.022 × 1023 atoms of chlorine.
(b) Under similar conditions of temperature and pressure, four 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 atom of an element is heavier than 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 molecules.
The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?
Answer:
According to Avogadro's law:
Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal numbers of molecules.
So, 1 mole of each gas contains = 6.02 × 1023 molecules
Mol. Mass of :
H2 = 2 g,
O2 = 32 g,
CO2 = 12 + 2(16) = 44 g,
SO2 = 32 + 2(16) = 64 g,
Cl2 = 2(35.5) = 71 g
(i) 2 g of hydrogen contains molecules = 6.02 × 1023
So, 8 g of hydrogen contains molecules
= x 8
= 4 × 6.02 × 1023molecules
= 24.08 × 1023molecules
(ii) 32 g of oxygen contains molecules = 6.02 × 1023
So, 8 g of oxygen contains molecules
= x 8
= 1.505 × 1023molecules
(iii) 44 g of carbon dioxide contains molecules = 6.02 × 1023
So, 8 g of carbon dioxide contains
= x 8
= 1.09 × 1023molecules
(iv) 64 g of sulphur dioxide contains molecules = 6.02 × 1023
So, 8g of sulphur dioxide contains
= x 8
= 0.75 × 1023molecules
(v) 71 g of chlorine contains molecules = 6.02 × 1023
So, 8g of chlorine contains
= x 8
= 0.67 × 1023molecules
Thus Cl2 < SO2 < CO2 < O2 < H2
(i) Least number of molecules in Cl2
(ii) Most number of molecules in H2
Numerical Problems
72 questionsAnswer:
(a) 504 g nitric acid reacts with 192 g of copper
∴ 63 g of nitric acid reacts with x 63 = 24 g of copper
Hence, 24 g of copper is required.
(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO
∴ 63 g of nitric acid gives x 63
= 5.6 litre of NO
5.6 L of NO is collected.
Answer:
(a) Gram molecular mass of N2 = 2 x 14 = 28 g
28 g of nitrogen = 1 mole
∴ 7 g of nitrogen = x 7
= 0.25 moles
(b) Gram molecular mass of Cl2 = 2 x 35.5 = 71 g
71 g of chlorine at S.T.P. occupies 22.4 litres
∴ 7.1 g of chlorine will occupy x 7.1
= 2.24 litre = 2.24 dm3
(c) Gram molecular mass of carbon monoxide (CO) = 12 + 16 = 28 g
28 g of carbon monoxide at S.T.P. occupies 22,400 cm3 volume
So, 22,400 cm3 volume have mass = 28 g
∴ 56 cm3 volume will have mass x 56 = 0.07 g
Answer:
(i) Molar mass of NaNO3 = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 g
Nitrogen in 85 g NaNO3 = 14 g
∴ Percentage of Nitrogen in NaNO3 = x 100 = 16.47%
(ii) Molar mass of (NH4)2SO4 = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 g
Nitrogen in 132 g of (NH4)2SO4 = 28 g
∴ Percentage of Nitrogen in (NH4)2SO4 = x 100 = 21.21%
(iii) Molar mass of CO(NH2)2 = 12 + 16 + 2[14 + (2 x 1)]
= 28 + 2(16)
= 28 + 32
= 60 g
Nitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g
∴ Percentage of Nitrogen in CO(NH2)2 = x 100 = 46.66%
So, the highest percentage of Nitrogen is in Urea.
Answer:
According to Gay lussac's law,
4 Vol of CO2 is collected with 2 Vol. of C2H2
So, 200 cm3 CO2 will be collected with
Similarly, 4 Vol of CO2 is produced by 5 Vol of O2
So, 200 cm3 CO2 will be produced by = x 200 = 250 cm3
Hence, acetylene = 100 cm3 and oxygen = 250 cm3
Answer:
Molar mass of urea [CO(NH2)2] = 12 + 16 + 2[14 + (2 x 1)]
= 28 + 2(16)
= 28 + 32
= 60 g
50 kg of urea = 50,000 g of urea
Nitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g
∴ Nitrogen in 50,000 g urea =
= 23,333 g = 23.3 kg
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Carbon | 80 | 12 | = 6.66 | = 1 |
Hydrogen | 20 | 1 | = 20 | = 3 |
Simplest ratio of whole numbers = C : H = 1 : 3
Hence, empirical formula is CH3
Empirical formula weight = 12 + 3(1) = 15
V.D. = 15
Molecular weight = 2 x V.D. = 2 x 15
∴ Molecular formula = n[E.F.] = 2[CH3] = C2H6
The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.
0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.
(a) Which elements does X contain?
(b) What was the purpose of copper (II) oxide?
(c ) Calculate the empirical formula of X by the following steps:
(i) Calculate the number of moles of carbon dioxide gas.
(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.
(iii) Calculate the mass of hydrogen in sample X.
(iv) Deduce the ratio of atoms of each element in X (empirical formula).
Answer:
(a) X contains carbon and hydrogen as it is a hydrocarbon.
(b) The purpose of copper (II) oxide is to act as an oxidizing agent.
(c) (i) 22400 cm3 CO2 has mass = 44 g
∴ 224 cm3 CO2 will have mass = x 224 = 0.44 g
Molar mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g
Moles of CO2 = = = 0.01 moles
(ii) Mass of carbon in 44 g CO2 = 12 g
∴ Mass of carbon in 0.44 g CO2 = x 0.44 = 0.12 g
As X contains carbon and hydrogen, so sample X has 0.12 g of carbon
(iii) Mass of Hydrogen in X = 0.145 - 0.12 = 0.025 g
(iv) Ratio of moles of C : H
= :
= 1 :
= 1 :
= 2 : 5
Hence, ratio of C : H = 2 : 5
so, the empirical formula of hydrocarbon is C2H5
Answer:
Element | Mass (g) | Atomic mass | Moles | Simplest ratio |
---|---|---|---|---|
Element X | 24 | 12 | = 2 | = 1 |
Oxygen | 64 | 16 | = 4 | = 2 |
Simplest ratio of whole numbers = X : O = 1 : 2
Hence, simplest formula of compound is XO2
Answer:
(a ) V.D. = = = 17
(b) Molecular weight = 2 x V.D. = 2 x 17 = 34 a.m.u.
(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :
CO2 + C ⟶ 2CO
What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?
(b) 60 cm3 of oxygen was added to 24 cm3 of carbon monoxide and mixture ignited. Calculate:
(i) volume of oxygen used up and
(ii) Volume of carbon dioxide formed.
Answer:
[By Gay Lussac's law]
1 Vol. of C produces 2V of CO
12 g of C produces 2 x 22.4 L of CO
∴ 3 g of C will produce x 3
= 11.2 L of CO
(b)
(i) 2 Vol. of CO requires 1 Vol. of oxygen
∴ 24 cm3 CO will require x 24
= 12 cm3 of oxygen
(ii) 2 Vol. of CO gives 2 vol. of CO2
∴ 24 cm3 of CO will give 24 cm3 of CO2
Answer:
328 g of calcium nitrate produces 112 g of CaO
∴ 82 g of calcium nitrate will produce x 82
= 28 g of CaO
328 g of calcium nitrate produces 4 x 22.4 L of NO2
∴ 82 g of calcium nitrate will produce x 82
= 22.4 L of NO2
The equation for the burning of octane is:
2C8H18 + 25O2 ⟶ 16CO2 + 18H2O
(i) How many moles of carbon dioxide are produced when one mole of octane burns?
(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?
(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?
(iv) What is the empirical formula of octane?
Answer:
(i) 2 moles of octane produces 16 moles of CO2
∴ 1 mole octane produces x 1
= 8 moles of CO2
(ii) 1 mole CO2 occupies volume = 22.4 L
As 2 moles of octane produces 8 moles of carbon dioxide which will occupy volume x 8
= 179.2 dm3 of CO2
(iii) 1 mole CO2 has mass = 44 g
∴ 16 moles will have mass x 16
= 704 g of CO2
(iv) Molecular formula is C8H18
∴ Ratio of C and H is 8 : 18
Simple ratio is 4 : 9
Hence, empirical formula = C4H9
Answer:
Element | Mass | At. wt. | gram atoms | Simplest ratio |
---|---|---|---|---|
Silicon | 5.6 | 28 | = 0.2 | = 1 |
Chlorine | 21.3 | 35.5 | = 0.6 | = 3 |
Simplest ratio of whole numbers = Si : Cl = 1 : 3
Hence, empirical formula is SiCl3
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Phosphorus | 38.27 | 31 | = 1.234 | = 1 |
Hydrogen | 2.47 | 1 | = 2.47 | = 2 |
Oxygen | 59.26 | 16 | = 3.70 | = 3 |
Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3
Hence, empirical formula is H2PO3
Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81
Molecular weight = 162
So, molecular formula = 2(H2PO3) = H4P2O6
Answer:
(a) Given, molecular mass of 'A' is 60
V1 = 10 L
T1 = 27 + 273 K = 300 K
P1 = 700 mm
T2 = 273 K
P2 = 760 mm
V2 = ?
Using formula:
=
Substituting in the formula,
=
V2 = = = 8.38 L
As, 22.4 L of A weighs 60 g
∴ 8.38 L of A will weigh x 8.38
= 22.446 = 22.45 g
(b) V1 = 360 cm3
T1 = 87 + 273 K = 360 K
P1 = 380 mm Hg pressure
T2 = 273 K
P2 = 760 mm Hg pressure
V2 = ?
Using formula:
=
Substituting in the formula,
=
V2 = = = 136.5 cm3
136.5 cm3 of gas weigh = 0.546 g
22400 cm3 of gas weigh x 22400
= 89.6 amu
A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.
(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?
(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result.
Answer:
(a) Given, cylinder can hold = 1 kg of hydrogen
Molar mass of hydrogen = 2[1] = 2g
Number of moles of hydrogen present in cylinder = = 500
According to avogadro's law : cylinder will hold 500 moles of carbon dioxide gas
Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g
1 mole of carbon dioxide = 44 g
∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.
Hence, Weight of carbon dioxide in cylinder = 22 kg
(b) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
∴ molecules in the cylinder of carbon dioxide = X
Avogadro's law helped to arrive at this result.
Following questions refer to one mole of chlorine gas.
(a) What is the volume occupied by this gas at S.T.P.?
(b) What will happen to the volume of gas, if pressure is doubled?
(c) What volume will it occupy at 273°C?
(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?
Answer:
(a) According to Avogadro's law : The volume occupied by 1 mole of chlorine is 22.4 dm3
(b) According to Boyle's Law: PV = constant
Hence, if pressure is doubled then volume will become half i.e. = 11.2 dm3
(c) V1 = 22.4 dm3
T1 = 273 K
T2 = 273 + 273 K = 546 K
V2 = ?
According to charles law:
=
Substituting to we get,
=
Hence, V2 = x 546 = 44.8 dm3
(c) Mass of 1 mole Cl2 gas = 35.5 × 2 = 71 g
(a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.
(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.
(c) How much volume will be occupied by 2g of dry oxygen at 27°C and 740 mm pressure?
(d) What would be the mass of CO2 occupying a volume of 44 litres at 25°C and 750 mm pressure?
(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.
AgNO3 (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO3
Calculate the percentage of NaCl in the mixture.
Answer:
Relative molecular mass of CuSO4.xH2O
= 40 + 32 + (4×16) + [x(2+16)]
= 40 + 32 + 64 + 18x
= 136 + 18x
∴ 21% water of crystallization = 18 x
Hence, water of crystallization = 2
(b) Molar mass of H2O = 2(1) + 16 = 18 g
For 18 g water, vol. of hydrogen needed = 22.4 litre
∴ For 1.8 g, vol. of H2 needed = x 1.8 = 2.24 L
According to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygen
When 2 vol of hydrogen in 1.8 g H2O is 2.24 L, then one vol. of oxygen will be:
= 1.12 L
(c) Gram molecular mass of O2 = 2 x 16 = 32 g
1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.
∴ 2 g of O2 occupies =
Volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure:
S.T.P. | Given Values |
---|---|
P1 = 760 mm of Hg | P2 = 740 mm of Hg |
V1 = 1.4 lit | V2 = x lit |
T1 = 273 K | T2 = 27 + 273 K |
Using the gas equation,
Substituting the values we get,
Hence, the volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure is 1.58 lit.
(d) Gram molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g
Given Values | S.T.P. |
---|---|
P1 = 750 mm of Hg | P2 = 760 mm of Hg |
V1 = 44 lit | V2 = x lit |
T1 = 25 + 273 K = 298 K | T2 = 273 K |
Using the gas equation,
Substituting the values we get,
22.4 litre of CO2 at S.T.P. has mass = 44 g
39.78 litre of CO2 at S.T.P. has mass x 39.78
= 78.14 g
(e)
(i) 143.5 g AgCl is formed by 58.5 g NaCl
∴ 1.435 g of AgCl will be formed by x 1.435 = 0.582 g
Percentage of NaCl = x 100 = 58.5%
Hence, percentage of NaCl is 58.5%
Answer:
(i) 196 g of sulphuric acid oxidizes 12 g carbon
∴ 49 g of sulphuric acid will oxidize = x 49 = 3 g
Hence, 3 g of carbon is oxidized.
(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO2
∴ 3 g of carbon will liberate x 3 = 11.2 lit of SO2.
Hence, 11.2 lit of SO2 is liberated.
(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)
(ii) The relative molecular mass of this compound is 168, so what is it's molecular formula?
Answer:
(i)
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Carbon | 14.4 | 12 | = 1.2 | = 1 |
Hydrogen | 1.2 | 1 | = 1.2 | = 1 |
chlorine | 84.5 | 35.5 | = 2.38 | = 1.98 = 2 |
Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2
Hence, empirical formula is CHCl2
Empirical formula weight = 12 + 1 + 2(35.5) = 84 g
Relative molecular mass = 168
Molecular formula = n[E.F.] = 2[CHCl2] = C2H2Cl4
∴ Molecular formula = C2H2Cl4
Answer:
Relative molecular mass of [Mg (NO3)2.6H2O]
= 24 + 2[14 + 3(16)] + 12(1) + 6(16)
= 24 + 2[14 + 48] + 12 + 96
= 24 + 28 + 96 + 12 + 96
= 256 g
Molar mass of oxygen in [Mg (NO3)2.6H2O] = 96 + 96 = 192 g
Since, 256 g of [Mg(NO3)2.6H2O] contains 192 g of oxygen
∴ 100 g of [Mg(NO3)2.6H2O] contains
x 100 = 75% of oxygen
(b) Relative molecular mass of Na2B4O7.10H2O
= 2(23) + 4(11) + 7(16) + 20(1) + 10(16)
= 46 + 44 + 112 + 20 + 160
= 382 g
Molar mass of boron in Na2B4O7.10H2O = 44 g
382 g of Na2B4O7.10H2O contains 44 g of boron
∴ 100 g Na2B4O7.10H2O contains x 100 = 11.5%
(c) Relative molecular mass of Ca(H2PO4)2
= 40 + 4(1) + 2(31) + 8(16)
= 40 + 4 + 62 + 128
= 234 g
Molar mass of phosphorus in Ca(H2PO4)2 = 62 g
234g of Ca(H2PO4)2 contains 62 g of phosphorus
∴ 100 g of Ca(H2PO4)2 contains x 100 = 26.5%
Answer:
80 g of sodium hydroxide precipitates 98 g of copper hydroxide.
∴ 200 g of sodium hydroxide will precipitate x 200 = 245 g of copper hydroxide.
Hence, 245 g. of copper hydroxide is precipitated.
Solid ammonium dichromate decomposes as under:
(NH₄)₂Cr₂O₇ ⟶ N₂ + Cr₂O₃ + 4H₂O
If 63 g of ammonium dichromate decomposes. Calculate
(a) the quantity in moles of (NH₄)₂Cr₂O₇
(b) the quantity in moles of nitrogen formed.
(c) the volume of N₂ evolved at S.T.P.
(d) the loss of mass
(e) the mass of chromium (III) oxide formed at the same time.
Answer:
(a) 252 g of (NH4)2Cr2O7 = 1 mole
∴ 63 g of (NH4)2Cr2O7 = x 63 = 0.25 moles
Hence, no. of moles = 0.25 moles
(b)
Hence, 0.25 moles of (NH4)2Cr2O7 will produce 0.25 moles of nitrogen.
(c) 1 mole of N₂ occupies 22.4 lit.
∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.
Hence, volume of N₂ evolved at s.t.p = 5.6 lit.
(d) 252 g of (NH4)2Cr2O7 decomposes to give 152 g of solid Cr₂O₃, loss in mass = 252 - 152 = 100 g
If 63 g of (NH4)2Cr2O7 decomposes then loss in mass is
x 63 = 25 g
(e) 1 mole of Cr2O7 = 152 g.
∴ 0.25 moles of Cr2O7 = 152 x 0.25 = 38 g.
Hence, mass in gms of Cr2O7 formed = 38 g.
Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:
2H2S + 3O2 ⟶ 2H2O + 2SO2
Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).
Answer:
128 g of SO2 has volume 2 × 22.4 litres
∴ 12.8 g of SO2 has volume =
x 12.8 = 4.48 L
Volume of oxygen = ?
2 × 22.4 L H2S requires = 3 × 22.4 litre of oxygen
∴ 4.48 L H2S will require
x 4.48 = 6.72 L of oxygen.
Answer:
When 60 g NO is formed then, 54 g mass of steam is produced
∴ when 1.5 g NO is formed, mass of steam produced
= x 1.5 = 1.35 g
Answer:
Molecular mass of Ca(NO3)2
= 40 + [2(14) + 6(16)]
= 40 + 28 + 96
= 164 g
Mass of N2 in Ca(NO3)2 = 28 g
In 1 hectare of soil, 20 kg of N2 is removed
∴ In 10 hectare field N2 removed is 20 x 10 = 200 kg
28 g N2 is present in 164 g of Ca(NO3)2
So, 200 kg or 200,000 g of N2 is present in x 200,000
= 1171428 g = 1171.4 kg
Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
P + 5HNO3 ⟶ H3PO4+ 5NO2 + H2O
If 6.2 g of phosphorus was used in the reaction calculate:
(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.
(b) mass of nitric acid consumed at the same time?
(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273°C?
Answer:
(a) 31 g of P = 1 mole
∴ 6.2 g of P = x 6.2 = 0.2 mole of P
Mass of phosphoric acid formed = ?
31 g of P produces 98 g of phosphoric acid
∴ 6.2 g of P will form x 6.2 = 19.6 g
(b) 31 g P reacts with 315 g HNO3
∴ 6.2 g P will react with x 6.2 = 63 g HNO3
(c) 31 g P produces = 1 mole steam
∴ 6.2 g P produces x 6.2 = 0.2 moles
Volume of steam produced at STP = 0.2 × 22.4 = 4.48 litre
V1 = 4.48 litre
T1 = 273 K
P1 = 760 mm Hg pressure
T2 = 273 + 273 = 546 K
P2 = 760 mm Hg pressure
V2 = ?
Using formula:
=
Substituting in the formula,
=
V2 = 2 x 4.48 = 8.96 L
Hence, volume of steam produced = 8.96 L
Answer:
Given,
112 cm3 of gaseous fluoride has mass = 0.63 g
so, 22400 cm3 of gaseous fluoride will have mass = x 22400
= 126 g
∴ Relative molecular mass of fluoride = 126 g
The molecular mass = At mass P + At. mass of F
126 = 31 + At. Mass of F
∴ At. Mass of F = 126 - 31 = 95 g
However, At. mass of F = 19
∴ = 5
So, 5 atoms of F, hence, the molecular formula = PF5
Answer:
286 g of washing soda had 106 g of anhydrous sodium carbonate
∴ 57.2 g will have = x 57.2 = 21.2 g anhydrous sodium carbonate.
Hence, 21.2 g anhydrous sodium carbonate is left.
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Metal M | 34.5 | 56 | = 0.616 | = 1 |
chlorine | 65.5 | 35.5 | = 1.84 | = 2.98 = 3 |
Simplest ratio of whole numbers = M : Cl = 1 : 3
Hence, empirical formula is MCl3
Empirical formula weight = 56 + 3(35.5) = 162.5 g
Molecular weight = 2 x V.D. = 2 x 162.5 = 325
Molecular formula = n[E.F.] = 2[MCl3] = M2Cl6
∴ Molecular formula = M2Cl6
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
carbon | 4.8 | 12 | = 0.4 | = 1 |
bromine | 95.2 | 80 | = 1.19 | = 2.9 = 3 |
Simplest ratio of whole numbers = C : Br = 1 : 3
Hence, empirical formula is CBr3
(ii) Empirical formula weight = 12 + 3(80) = 252 g
Molecular weight = 2 x V.D. = 2 x 252 = 504
Molecular formula = n[E.F.] = 2[CBr3] = C2Br6
∴ Molecular formula = C2Br6
Answer:
2 x 22400 litre steam is produced by 4 × 22400 cm3 N2O
∴ 150 cm3 steam will be produced by
= x 150 = 300 cc of N2O
Samples of the gases O2, N2, CO2 and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure
What is the volume occupied by:
(a) x molecules of N2
(b) 3x molecules of CO
(c) What is the mass of CO2 in grams?
(d) In answering the above questions, which law have you used?
Answer:
By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.
∴ If gases under the same conditions have same number of molecules then they must have the same volume.
(i) So, X molecules of N2 occupy 1V litres.
(ii) 3X molecules of CO will occupies 3V litres.
(iii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g
1 mole of O2 weighs 32 g and occupy 22.4 lit. volume
∴ 8 g of O2 will occupy x 8 = 5.6 lit. vol. = Molar volume
If 8 g of O2 occupies 5.6 lit. vol.
And 1 mole of CO2 occupy 22.4 lit and weighs = 44 g at s.t.p.
∴ 5.6 lit. vol. of CO2 will weigh = x 5.6 = 11 g
Hence, mass of CO2 = 11 g
(iv) Avogadro's Law is used above.
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
sodium | 42.1 | 23 | = 1.83 | = 3 |
phosphorus | 18.9 | 31 | = 0.609 | = 1 |
oxygen | 39 | 16 | = 2.43 | = 4 |
Simplest ratio of whole numbers = Na : P : O = 3 : 1 : 4
Hence, empirical formula is Na3PO4
Answer:
From equation:
22.4 dm3 of methane requires oxygen = 2 x 22.4 dm3 of O2 = 44.8 dm3
From equation,
[2 x 22.4] dm3 hydrogen requires oxygen = 22.4 dm3
∴ 11.2 dm3 hydrogen will require oxygen = x 11.2 =
= 5.6 dm3
Total volume of oxygen required = 44.8 + 5.6 = 50.4 dm3
Answer:
Element | Mass (g) | Atomic mass | Moles | Simplest ratio |
---|---|---|---|---|
Pb | 6.21 | 207 | = 0.03 | = 1 |
Cl | 4.26 | 35.5 | = 0.12 | = 4 |
Simplest ratio of whole numbers Pb : Cl = 1 : 4
Hence, empirical formula is PbCl4
10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:
Na2SO4 + BaCl2 ⟶ BaSO4 + 2NaCl
Calculate the percentage of sodium sulphate in the original mixture.
Answer:
233 g of BaSO4 is obtained from 142 g of Na2SO4
6.99 g of BaSO4 will be obtained from x 6.99 = 4.26 g of Na2SO4.
∴ In 10 g mixture x 100 = 42.6% Na2SO4 is present.
Hence, 42.6% Na2SO4 is present in the original mixture
When heated, potassium permanganate decomposes according to the following equation:
2KMnO4 ⟶ K2MnO4 + MnO2 + O2
(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)
Answer:
2KMnO4 ⟶ K2MnO4 + MnO2 + O2
Loss in mass = 1.32 g = 1 lit of oxygen
Vapour density of gas =
Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g
Hence, relative molecular mass of oxygen is 32 g
(b) Molar mass of 2KMnO4 = 2[39 + 55 + 4(16)] = 2[39 + 55 + 64] = 316 g
316 g of KMnO4 gives oxygen = 24 litres
∴ 15.8 g of KMnO4 will give
= x 15.8 = 1.2 L
Answer:
(a) Molar mass of sulphur dioxide = 32 + 2(16) = 64 g
64 g of sulphur dioxide = 1 mole
So, 3.2 g = x 3.2 = 0.05 moles
(b) 1 mole of SO2 = 6.02 × 1023 molecules
So, in 0.05 moles, no. of molecules = 6.02 × 1023 × 0.05 = 3 × 1022
(c) The volume occupied by 64 g of SO2 = 22.4 dm3
3.2 g of SO2 will be occupied by volume
x 3.2 = 1.12 L
The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure .
(i) Which sample of gas contains the maximum number of molecules?
(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?
(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?
(iv) If the volume of A is actually 5.6 dm3 at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro's number is 6 × 1023).
(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O)
Answer:
(i) Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.
(ii) If number of molecules of gas A is doubled, the volume will also be doubled i.e. 2V.
(iii) Gay Lussac's Law is observed.
(iv) 1 mole contains 6 x 1023 number of molecules and occupies 22.4 lit. vol.
Given, volume of 'A' is 5.6 dm3 at s.t.p.
∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 1023 (Avogadro no.)
(v) As D is 1 mole hence, mass of 1 mole of D (N2O) = 2(14) + 16 = 28 + 16 = 44 g
Hence, mass of N2O = 44 g
The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2CO3 = 106]
(i) NaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4Cl
(ii) 2NaHCO3 ⟶ Na2CO3 + H2O + CO2
Equations (1) and (2) are based on the production of 21.2 g. of sodium carbonate.
(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate
(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.
Answer:
106 g of Na2CO3 is obtained from 168 g of NaHCO3
∴ 21.2 g. of Na2CO3 is obtained from x 21.2 = 33.6 g of NaHCO3
(ii)
84 g of NaHCO3 is obtained from 22.4 lit of CO2
∴ 33.6 g. of NaHCO3 is obtained from x 33.6 = 8.96 lit.
Hence, 8.96 lit of CO2 is required.
A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)
NH4NO3 ⟶ N2O + 2H2O
(i) What volume of di nitrogen oxide is produced at the same time as 8.96 litres of steam.
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam [Relative molecular mass of NH4NO3 is 80]
(iii) Determine the percentage of oxygen in ammonium nitrate [O = 16]
Answer:
(i) Given,
1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)
Hence, Vol of di nitrogen = = 4.48 lit.
Hence, volume of di nitrogen oxide produced = 4.48 lit.
(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH4NO3
∴ 8.96 lit of steam will be produced by x 8.96 = 16 g
Hence, 16 g of ammonium nitrate is required to be heated.
(iii) % of oxygen in ammonium nitrate = x 100 = 60 %
Answer:
240 g of CuO is reduced by 44.8 lit of NH3
∴ 120 g of CuO is reduced by x 120 = 22.4 lit.
Hence, 22.4 lit of NH3 is required
Answer:
Gram molecular mass of C2H4
= 2(12) + 4(1)
= 24 + 4 = 28 g
As,
28 g of C2H4 = 1 mole
∴ 1.4 g of C2H4 = x 1.4 = 0.05 moles
1 mole = 6 × 1023 molecules
∴ 0.05 moles = 6 × 1023 x 0.05 = 3 x 1022 molecules
Hence, no. of moles is 0.05 and no. of molecules is 3 x 1022.
Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.
Hence, vol. occupied is 1.12 lit
(b)
Hence, vapour density is 14
(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) correct to the nearest whole number.
(F = 19; Na =23; Al = 27)
(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows :
2CO + O2 ⟶ 2CO2
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Answer:
Molecular weight of sodium aluminium fluoride (Na3AlF6)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g
210 g of sodium aluminium fluoride contains 69 g of Na
∴ 100 g of sodium aluminium fluoride will contain = x 100 = 32.85% = 33%
Hence, percentage of sodium in sodium aluminium fluoride (Na3AlF6) is 33%
(b)
1 mole of O2 has volume = 22400 ml
Volume of oxygen used by 2 × 22400 ml CO = 22400 ml
∴ Vol. of O2 used by 560 ml CO
= x 560
= 280 ml
Volume of CO2 formed by 2 × 22400 ml CO = 2 x 22400 ml
∴ Vol. of CO2 formed by by 560 ml CO
= 560 ml
Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.
2C2H2(g) + 5O2(g) ⟶4CO2(g) + 2H2O(g)
Answer:
According to Gay-Lussac's law,
2 volume of acetylene requires 5 volume of oxygen
1 volume of acetylene requires = 2.5 volume of oxygen
∴ 200 cm3 of acetylene requires x 200
= 500 cm3 of oxygen
2 volume of acetylene produces 4 volume of CO2
1 volume of acetylene produces = 2 volume of CO2
∴ 200 cm3 produces x 200 = 400 cm3 of CO2
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Nitrogen | 100 - 12.5 = 87.5 | 14 | = 6.25 | = 1 |
Hydrogen | 12.5 | 1 | = 12.5 | = 2 |
Simplest ratio of whole numbers N : H = 1 : 2
Hence, empirical formula is NH2
Empirical formula weight = 14 + 2(1) = 16
Relative molecular mass = 37
∴ Molecular formula = n[E.F.] = 2[NH2] = N2H4
Answer:
(1) Gram molecular mass of N2 = 2(14) = 28 g
6 × 1023 molecules of N2 weighs 28 g
∴ 24 × 1024 molecules will weigh x 24 × 1024 = 1120 g
(2) 6 × 1023 molecules of N2 occupies 22.4 dm3
∴ 24 × 1024 molecules of N2 will occupy x 24 × 1024 = 896 dm3
Commercial sodium hydroxide weighing 30g has some sodium chloride in it. The Mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is
NaCl + AgNO3 ⟶ AgCl + NaNO3
(Relative molecular mass of NaCl = 58; AgCl = 143)
Answer:
(i) 143 g AgCl is formed by 58 g NaCl
∴ 14.3 g of AgCl will be formed by x 14.3 = 5.8 g
Percentage of NaCl = x 100 = 19.33%
Hence, percentage of NaOH is 19.33%
LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:
C3H8 + 5O2 ⟶ 3CO2 + 4H2O
2C4H10 + 13O2 ⟶ 8CO2 + 10H2O
Answer:
Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres
1 Vol. C3H8 produces carbon dioxide = 3 Vol
So, 6 litres C3H8 will produce carbon dioxide = 3 x 6 = 18 litres
2 Vol. C4H10 produces carbon dioxide = 8 Vol
So, 4 litres C4H10 will produce carbon dioxide = x 4 = 16 litres
Hence, 34 (i.e., 18 + 16) litres of CO2 is produced.
Answer:
Given,
molecular mass of ammonium nitrate = 80
mass of nitrogen in 1 mole of ammonium nitrate (NH4NO3) is 2(14) = 28 g
percentage of nitrogen in ammonium nitrate = x 100 = 35%
mass of oxygen in 1 mole of ammonium nitrate (NH4NO3) is 3(16) = 48 g
percentage of oxygen in ammonium nitrate = x 100 = 60%
Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%
4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.
(i) Write the equation for the reaction.
(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)
(iii) What is the volume of carbon dioxide liberated at STP?
(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).
(v) How many moles of HCl are used in this reaction?
Answer:
(i) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2
(ii) Given,
1 mole of CaCO3 = molecular mass of CaCO3 = 100 g
∴ 4.5 moles of CaCO3 weighs x 4.5 = 450 g
(iii) 1 mole of CaCO3 produces 1 mole of CO2 and 1 mole occupies 22.4 l of volume.
∴ 4.5 moles of CaCO3 will produce 4.5 moles of CO2 and 4.5 moles will occupy 22.4 x 4.5 = 100.8 L
(iv) 1 mole CaCO3 produces 111 g. CaCl2
∴ 4.5 moles of CaCO3 will produce = 111 x 4.5 = 499.5 g.
(v)
∴ number of moles of HCl used = 2 x 4.5 = 9 moles.
Hence, moles of HCl used = 9 moles.
i. Calculate the volume of 320 g of SO2 at STP. (Atomic mass: S = 32 and O = 16).
ii. State Gay-Lussac's Law of combining volumes
iii. Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8). (Atomic mass: C = 12, O = 16, H = I, Molar Volume = 22.4 dm3 at stp.)
Answer:
(i) Gram molecular mass of SO2 = 32 + 2(16) = 32 + 32 = 64 g
64 g of SO2 occupy 22.4 lit of vol.
320 g of SO2 will occupy = x 320 = 112 lit.
Hence, volume of 320 g of SO2 = 112 lit.
(ii) Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."
(iii)
(i) 44 g propane requires 5 x 22.4 lit of oxygen
∴ 8.8 g of propane will require x 8.8 = 22.4 lit.
Hence, 22.4 lit of Oxygen is required.
Answer:
Element | % composition | At. wt. | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
C | 12.67 | 12 | = 1.05 | = 1 |
H | 2.13 | 1 | = 2.13 | = 2.02 = 2 |
Br | 85.11 | 80 | = 1.06 | = 1 |
Simplest ratio of whole numbers = C : H : Br = 1 : 2 : 1
Hence, empirical formula is CH2Br
Empirical formula weight = 12 + 2(1) + 80 = 94
Vapour density (V.D.) = 94
Molecular weight = 2 x V.D. = 2 x 94
∴ Molecular formula = n[E.F.] = 2[CH2Br] = C2H4Br2
Answer:
(i) gram molecular mass of S = 32
6 × 1023 atoms weigh = 32 g
1022 atoms will weigh = x 1022 = 0.533 g
(ii) 1 mole of carbon dioxide weighs = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g
∴ 0.1 mole of carbon dioxide weighs = 44 x 0.1 = 4.4 g.
Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:
P + 5HNO3 [conc.] ⟶ H3PO4 + H2O + 5NO2.
If 9.3 g of phosphorus was used in the reaction, calculate :
(i) Number of moles of phosphorus taken.
(ii) The mass of phosphoric acid formed.
(iii) The volume of nitrogen dioxide produced at STP.
Answer:
(i)
31 g of P = 1 mole
∴ 9.3 g of P = x 9.3 = 0.3 moles.
Hence, 0.3 moles of phosphorous was taken for the reaction.
(ii)
31 g of P forms 98 g of phosphoric acid
∴ 9.3 g will form x 9.3 = 29.4 g.
Hence, 29.4 g. of phosphoric acid is formed
(iii)
31 g of P produces 5 vol = 5 x 22.4 lit.
∴ 9.3 g will produce = x 9.3 = 33.6 lit.
Answer:
[By Lussac's law]
To calculate the volume of ammonia gas formed.
Hence, volume of NH3 formed is 44.8 lit.
We know,
∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.
Hence, 22.4 lit of nitrogen remains in the resultant mixture.
Answer:
Molecular weight of Mg(NO3)2.6H2O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6[18]
= 24 + 2(62) + 108
= 24 + 124 + 108 = 256 g
256 g of magnesium nitrate crystals contains 24 g of magnesium
∴ 100 g of magnesium nitrate crystals will contain = x 100 = 9.375% = 9.38%
Hence, percentage of magnesium in magnesium nitrate crystals is 9.38%
Answer:
Given, V.D. = 8
Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.
16 g occupies 22.4 lit.
∴ 24 g. will occupy = x 24 = 33.6 lit.
Hence, volume occupied by gas = 33.6 lit.
Answer:
According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.
Hence, number of molecules of N2 = Number of molecules of H2 = X
O2 is evolved by heating KClO3 using MnO2 as a catalyst.
2KClO3 2KCl + 3O2
(i) Calculate the mass of KClO3 required to produce 6.72 litre of O2 at STP.
(atomic masses of K = 39, Cl = 35.5, 0 = 16).
(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.
(iii) Calculate the volume occupied by 0.01 mole of CO2 at STP.
Answer:
(i)
67.2 lit. of O2 is produced by 245 g of KClO3
∴ 6.72 lit of O2 will be obtained from x 6.72 = 24.5 g
(ii)
22.4 lit = 1 mole
∴ 6.72 lit = x 6.72 = 0.3 moles
1 mole = 6.023 x 1023 molecules
∴ 0.3 moles = 0.3 x 6.023 x 1023 = 1.806 x 1023 molecules.
(iii)
Volume occupied by 1 mole of CO2 at STP = 22.4 litres
So, volume occupied by 0.01 mole of CO2 at STP
= 22.4 × 0.01
= 0.224 litres
Answer:
[By Lussac's law]
To calculate the volume of ethyne gas
∴
Hence, volume of ethyne gas required = 4.2 dm3.
Answer:
Empirical formula is X2Y
Empirical formula weight = 2(10) + 5 = 25
Vapour density (V.D.) = 25
Molecular weight = 2 x V.D. = 2 x 25
∴ Molecular formula = n[E.F.] = 2[X2Y] = X4Y2
Answer:
By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.
(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.
17 g. occupies 22.4 lit. of vol.
∴ 68 g will occupy = x 68 = 89.6 lit.
Hence, volume occupied by this gas = 89.6 lit.
(ii) 17 g = 1 mole
∴ 68 g = x 68 = 4 moles.
1 mole = 6 × 1023
∴ 4 moles = 4 x 6.023 × 1023 molecules.
Hence, Moles = 4
(iii) molecules = 4 x 6.023 × 1023 = 2.4 x 1024 molecules
Answer:
Element | Mass (g) | Atomic mass | Moles | Simplest ratio |
---|---|---|---|---|
Pb | 6.21 | 207 | = 0.03 | = 1 |
Cl | 4.26 | 35.5 | = 0.12 | = 4 |
Simplest ratio of whole numbers Pb : Cl = 1 : 4
Hence, empirical formula is PbCl4
Answer:
Element | % composition | Atomic weight | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
C | 12.67 | 12 | = 1.05 | = 1 |
H | 2.13 | 1 | = 2.13 | = 2.02=2 |
Br | 85.11 | 80 | = 1.06 | = 1 |
Simplest ratio of whole numbers = C : H : Br = 1 : 2 : 1
Hence, empirical formula is CH2Br
Empirical formula weight = 12 + 2(1) + 80 = 94
Vapour density (V.D.) = 94
Molecular weight = 2 x V.D. = 2 x 94
∴ Molecular formula = n[E.F.] = 2[CH2Br] = C2H4Br2
Answer:
Element | % composition | Atomic weight | Relative no. of atoms | Simplest ratio |
---|---|---|---|---|
Carbon | 4.8 | 12 | = 0.4 | = 1 |
Bromine | 95.2 | 80 | = 1.19 | = 2.9=3 |
Simplest ratio of whole numbers = C : Br = 1 : 3
Hence, empirical formula is CBr3
(i) Empirical formula weight = 12 + 3(80) = 252 g
Molecular weight = 2 x V.D. = 2 x 252 = 504
Molecular formula = n[E.F.] = 2[CBr3] = C2Br6
∴ Molecular formula = C2Br6