Mole Concept and Stoichiometry Solutions for Class 10 Science ICSE Board (Concise - Selina Publishers)
Free download of step by step solutions for class 10 Science (Chemistry) Chapter 5 - Mole Concept and Stoichiometry of ICSE Board (Concise - Selina Publishers). All exercise questions are solved & explained by an expert teacher and as per ICSE board guidelines.
Access ICSE Selina Solutions for Grade 10 Chemistry Chapter 5. - Mole Concept and Stoichiometry
1. State:
(a) Gay-Lussac's Law of combining volumes.
Ans: Gay-Lussac's law states that 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
Ans: Avogadro's law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
2. (a) What do you mean by stoichiometry?
Ans: Stoichiometry determines the amount of products/reactants produced/needed in a given reaction by measuring quantitative relationships.
(b) Define atomicity of a gas. State the atomicity of Hydrogen, Phosphorus and Sulphur.
Ans: Atomicity refers to the no of atoms present in a molecule. Atomicity of hydrogen is 2, phosphorus is 4 and sulphur is 8.
(c) Differentiate between N2 and 2N.
Ans:
N2 | 2N |
It consists of one molecule of nitrogen. | It consists of two atoms of nitrogen. |
Exist independently. | Cannot exist independently. |
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."
Ans: This is due to Avogadro's Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas = volume of helium gas
n molecules of hydrogen = n molecules of helium gas
nH2 = nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen are double the atoms of helium.
(b) "When stating the volume of a gas, the pressure and temperature should also be given."
Ans: For a given volume of gas under given temperature and pressure, a change in any one of the variables i.e., pressure or temperature changes the volume.
(c) Inflating a balloon seems to violate Boyle's law.
Ans: Inflating a balloon seems violating Boyle's law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.
4. (a) Calculate the volume of oxygen at S.T.P required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.
2CO + O2 $\to ~$2CO2
Ans: 2CO + O2 $\to $ 2CO2
2V 1V 2V
2 V of CO requires = 1V of O2
so, 100 litres of CO requires = 50 litres of O2
(b) 200 cm3 of hydrogen and 150 cm3 of oxygen are mixed and ignited, as per the following reaction,
2H2 + O2 $\to $2H2O
What volume of oxygen remains unreacted?
Ans: 2H2 + O2 $\to $2H2O
2V 1V 2V
From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm3 of Hydrogen reacts with = $\frac{200}{2}$= 100 cm3
Hence, the unreacted oxygen is 150 - 100 = 50cm3 of oxygen.
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 calculation.
Ans: This experiment supports Gay lussac's law of combining volumes.
Since the unchanged or remaining O2 is 58 cc so, used oxygen 106 - 58 = 48cc
According to Gay lussac's law, the volumes of gases reacting should be in a simple ratio.
CH4 + 2O2$\to$CO2 + 2H2O
1V 2V
24cc 48 cc
i.e. methane and oxygen react in a 1:2 ratio.
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 $\to $4CO2 + 2H2O (l)
Ans: 2C2H2 + 5O2 $\to $ 4CO2 + 2H2O (l)
2V 5V 4V
From equation, 2 V of C2H2 requires = 5 V of O2
So, for 400ml C2H2, O2 required = 400 \[\times\] $\frac{~~5~~}{2}$ =1000 ml
Similarly, 2 V of C2H2 gives = 4 V of CO2
So, 400ml of C2H2 gives CO2 = 400 ×$\frac{~4}{2}$ = 800ml
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
Ans: H2S + Cl2 $\to$ 2HCl + S
and calculate
(i) the volume of gaseous product formed
Ans: At STP, 1 mole gas occupies 22.4 L.
As 1 mole H2S gas produces 2 moles HCl gas,
22.4 L H2S gas produces 22.4 × 2 = 44.8 L HCl gas.
Hence, 112 cm3 H2S gas will produce 112 × 2 = 224 cm3 HCl gas.
(ii) composition of the resulting mixture.
Ans: 1 mole H2S gas consumes 1 mole Cl2 gas.
This means 22.4 L H2S gas consumes 22.4 L Cl2 gas at STP.
Hence, 112 cm3 H2S gas consumes 112 cm3 Cl2 gas.
120 cm3 - 112 cm3 = 8 cm3 Cl2 gas remains unreacted.
Thus, the composition of the resulting mixture is 224 cm3HCl gas + 8 cm3 Cl2 gas.
8. 1250cc of oxygen was burnt with 300cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed:
2C2H6+7O2 $\to $ 4CO2+6H2O
Ans: From the equation, 2V of ethane reacts with 7V oxygen.
So, 300 cc of ethane reacts with $\frac{~~~300\times 7~~}{2}$=1050cc
Hence, unused O2 = 1250 - 1050 = 200 cc
From 2V of ethane, 4V of CO2 is produced.
So, 300 cc of ethane will produce $\frac{~~~300\times 4~~}{2}$=600cc of CO2
9. What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C2H4] at 273o C and 380 mm of Hg pressure?
C2H4+3O2$\to $2CO2 + 2H2O
Ans:
C2H4+3O2$\to $2CO2 + 2H2O
1V 3V
$\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{1}}}}$= $\frac{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{2}}}}$
V2= \[\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}{{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{1}}}}\]=$\frac{~~380\times 33\times 273~~}{549\times 760}$=8.25 litres
10. Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at S.T.P.
CH4 + 2Cl2$\to $CH2Cl2 + 2HCl
Ans: CH4 + 2Cl2$\to $ CH2Cl2 +2HCl
1V 2V V 2V
From equation, 1V of CH4 gives = 2 V HCl
so, 40 ml of methane gives = 80 ml HCl
For 1V of methane = 2V of Cl2 required
So, for 40ml of methane = 40 x 2 = 80 ml of Cl2
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$\to $3CO2 + 4H2O
Ans: C3H8 + 5O2\[\rightarrow\] 3CO2 + 4H2O
1V 5V 3V
From equation, 5V of O2 required = 1V of propane
so, 100 cm3 of O2 will require = 20 cm3 of propane
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$\to $2NO2
Ans: 2NO + O2$\to $2NO2
2V V 2V
From equation, 1V of O2 reacts with = 2V of NO
200cm3 oxygen will react with = 200 $\times $2 =400 cm3 NO
Hence, remaining NO is 450 - 400 = 50 cm3
NO2 produced = 400cm3 because 1V oxygen gives 2V NO2
Total mixture = 400 + 50 = 450 cm3
13. If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.
Ans: 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.
When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leaving behind only 2 litres of hydrogen gas.
Therefore, the volume of the residual gas will be 2 litres.
14. Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.
4NH3 + 5O2$\to $4NO + 6H2O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?
Ans: 4NH3 + 5O2$\to $4NO + 6H2O
4V 5V 4V
9 litres of reactants gives 4 litres of NO
So, 27 litres of reactants will give = 27 \[\times \frac{~~4~~}{9}\] = 12 litres of NO
15. A mixture of hydrogen and chlorine occupying 36 cm3 exploded. On shaking it with water, 4cm3 of hydrogen was left behind. Find the composition of the mixture.
Ans: H2 + Cl2 $\to $2HCl
1V 1V 2V
Since 1V hydrogen requires 1V of oxygen and 4cm3 of H2 remained behind so the mixture had composition: 16 cm3 hydrogen and 16 cm3 chlorine.
Therefore Resulting mixture is H2 = 4cm3, HCl = 32cm3
16. What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 each of methane and acetylene?
CH4 + 2O2 $\to $CO2 + 2H2O
2C2H2 + 5O2$\to $4CO2 + 2H2O
Ans: CH4 + 2O2 $\to $CO2 + 2H2O
1V 2V 1V
2C2H2 + 5O2$\to $4CO2 + 2H2O
2V 5V 4V
From the equations, we can see that
1V CH4 requires oxygen = 2V O2
So, 10cm3 CH4 will require =20 cm3 O2
Similarly 2V C2H2 requires = 5V O2
So, 10 cm3 C2H2 will require= 25 cm3 O2
Now, 20 V O2 will be present in 100 V air and 25 V O2 will be present in 125 V air ,so the volume of air required is 225cm3
17. LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to the atmosphere.
C3H8 + 5O2$\to $3CO2 + 4H2O
2C4H10 + 13O2$\to$8CO2 + 10H2O
Ans: C3H8 + 5O2$\to $3CO2 + 4H2O
2C4H10 + 13O2$\to $8CO2 + 10H2O
60 ml of propane (C3H8) gives 3 $\times $60 = 180 ml CO2
40 ml of butane (C4H10) gives = 8 $\times $ \[\frac{40}{2}\] = 160 ml of CO2
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO2 is produced.
18. 200 cm3 of CO2 is collected at S.T.P when a mixture of acetylene and oxygen is ignited. Calculate the volume of acetylene and oxygen at S.T.P. in the original mixture.
2C2H2(g) + 5O2(g) $\to $4CO2(g)+ 2H2O(g)
Ans: 2C2H2(g) + 5O2(g) $\to $4CO2(g)+ 2H2O(g)
4 V CO2 is collected with 2 V C2H2
So, 200cm3 CO2 will be collected with = 100cm3 C2H2
Similarly, 4V of CO2 is produced by 5 V of O2
So, 200cm3 CO2 will be produced by = 250 ml of O2
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 litres 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: According to Avogadro's law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and less volume will contain less molecules.
(a) 5 litres of hydrogen has the greatest no. of molecules with the maximum volume.
(b) 1 litre of SO2 contains the least number of molecules since it has the smallest volume.
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.
Gas | Volume (in litres) | Number of molecules |
Chlorine Nitrogen Ammonia Sulphur dioxide | 10 20 20 5 | x |
Ans:
Gas | Volume (in litres) | Number of molecules |
Chlorine | 10 | \[\frac{x}{2}\] |
Nitrogen | 20 | x |
Ammonia | 20 | x |
Sulphur dioxide | 5 | \[\frac{x}{4}\] |
21. (i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B?
The gases A and B are under the same conditions of temperature and pressure.
Ans: According to Avogadro's law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
As 150 cc of gas A contains X molecules, 150 cc of gas B also contains X molecules.
So, 75 cc of B will contain X/2 molecules.
(ii) Name the law on which the above problem is based.
Ans: The problem is based on Avogadro's law.
Exercise – 5B
1. (a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.
Ans: This statement states that one atom of chlorine is 35.5 times heavier than 1/12 times of the mass of an atom C-12.
(b) What is the value of Avogadro's number?
Ans: 6.023 $\times $1023
(c) What is the value of molar volume of a gas at S.T.P?
Ans: The molar volume of a gas at STP is 22.4 dm3 at STP.
2. Define or explain the terms
(a) Vapour density
Ans: The ratio between the masses of equal quantities of gas and hydrogen at normal temperature and pressure is known as the vapour density.
(b) Molar volume
Ans: Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm3.
(c) Relative atomic mass
Ans: The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.
(d) Relative molecular mass
Ans: The relative molecular mass of a compound is the number that represents how many times one molecule of the substance is heavier than 1/12 of the mass of an atom of carbon-12.
(e) Avogadro's number
Ans: The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x1023 atoms.
(f) Gram atom
Ans: The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.
(g) Mole
Ans: Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.
3. (a) What are the main applications of Avogadro's Law?
Ans: (a) Applications of Avogadro's Law:
(1) It explains Gay-Lussac's law.
(2) It determines the atomicity of the gases.
(3) It determines the molecular formula of a gas.
(4) It determines the relation between molecular mass and vapour density.
(5) It gives the relationship between gram molecular mass and gram molecular volume.
(b) How dose Avogadro's Law explain Gay-Lussac's Law of combining volumes?
Ans: 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 what Gay Lussac's Law says.
H2 + Cl2 $\to $ 2HCl
1V 1V 2V(By Gay-Lussacs law)
4. Calculate the relative molecular masses of:
(a) Ammonium chloroplatinate, (NH4)2 PtCl6
Ans: (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) Potassium chlorate
Ans: KClO3
= (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) CuSO4. 5H2O
Ans: (Cu)63.5 + (S)32 + (4O)64 + (5H2O)5 x 18 = 249.5
(d) (NH4)2SO4
Ans: (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) CH3COONa
Ans: (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) CHCl3
Ans: (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (NH4)2 Cr2O7
Ans: (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252
5. Find the
(a) number of molecules in 73 g of HCl,
Ans: No. of molecules in 73 g HCl
= 6.023 $\times $1023$\times $ 73/36.5(mol. mass of HCl)
= 12.04 $\times $ 1023
(b) weight of 0.5 mole of O2,
Ans: Weight of 0.5 mole of O2 is
= 32(mol. Mass of O2) $\times $ 0.5
=16 g
(c) number of molecules in 1.8 g of H2O
Ans: No. of molecules in 1.8 g H2O
= 6.023 $\times $ 1023 $\times \frac{~~~18~~}{18}$
= 6.023 $\times $ 1022
(d) number of moles in 10 g of CaCO3
Ans: No. of moles in 10g of CaCO3
= $\frac{~~~10~~}{100}$(mol. Mass CaCO3)
= 0.1 mole
(e) Weight of 0.2 mole of H2 gas,
Ans: Weight of 0.2 mole H2 gas
= 2(Mol. Mass) $\times $ 0.2
= 0.4 g
(f) Number of molecules in 3.2 g of SO2.
Ans: No. of molecules in 3.2 g of SO2
= 6.023 × 1023 × \[\frac{3.2}{64}\]
= 3.023 × 1022
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
Weight of H2O= 2+16=18
Weight of CO2= 12+32=44
Weight of NH3= 14+3=17
Weight of CO= 12+16=28
7. Which of the following contains the 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: (b) 4 g of NH3
4g of NH3 having minimum molecular mass contains maximum molecules.
8. Calculate the number of
(a) Particles in 0.1 mole of any substance.
Ans: No. of particles in s1 mole = 6.023 $\times $ 1023
So, particles in 0.1 mole = 6.023 $\times $ 1023 $\times $ 0.1
= 6.023 $\times $ 1022
(b) Hydrogen atoms in 0.1 mole of H2SO4.
Ans: 1 mole of H2SO4 contains =2 $\times $ 6.023 $\times $1023
So, 0.1 mole of H2SO4 contains =2 $\times $6.023 $\times $ 1023 $\times $ 0.1
= 1.2$\times $1023 atoms of hydrogen
(c) Molecules in one Kg of calcium chloride.
Ans: 111g CaCl2 contains = 6.023 $\times $ 1023 molecules
So, 1000 g contains = 5.42 $\times $ 1024 molecules
9. How many grams of
(a) Al are present in 0.2 mole of it?
Ans: 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 $\times $ 27 = 5.4 g
(b) HCl are present in 0.1 mole of it?
Ans: 0.1 mole of HCl has mass
=0.1 $\times $ 36.5(mass of 1 mole)
= 3.65 g
(c) H2O are present in 0.2 mole of it?
Ans: 0.2 mole of H2O has mass = 0.2 $\times $ 18 = 3.6 g
(d) CO2 is present in 0.1 mole of it?
Ans: 0.1 mole of CO2 has mass = 0.1 $\times $ 44 = 4.4 g
10. (a) The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas?
Ans: 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 $\times $ \[\frac{22.4}{5.6}\]
= 48 g(molar mass)
(b) Calculate the volume occupied at S.T.P. by 2 moles of SO2.
Ans: 1 mole of SO2 has volume = 22.4 litres
So, 2 moles will have = 22.4 $\times $ 2 = 44.8 litre
11. Calculate the number of moles of
(a) CO2 which contain 8.00 g of O2
Ans: 1 mole of CO2 contains O2 = 32g
So, CO2 having 8 gm of O2 has no. of moles = $\frac{~~8~~}{32}$
= 0.25 moles
(b) Methane in 0.80 g of methane.
Ans: 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = $\frac{~0.8}{16}$
= 0.05 moles
12. Calculate the actual mass of
(a) An atom of oxygen
Ans: 6.023 $\times $ 10 23 atoms of oxygen has mass = 16 g
So, 1 atom has mass = $\frac{~~~~16~~}{6.023}\times $ 1023
= 2.656 $\times $ 10-23 g
(b) an atom of hydrogen
Ans: 1 atom of Hydrogen has mass = $\frac{~~~~1.6~~}{6.023}\times $ 1023 = 1.666 $\times $ 10-24
(c) a molecule of NH3
Ans: 1 molecule of NH3 has mass = $\frac{~~~~17~~}{6.023}\times $ 1023 = 2.82 $\times $ 10-23 g
(d) the atom of silver
Ans: 1 atom of silver has mass = $\frac{~~~~108~~}{6.023}\times $ 1023 =1.701 $\times $ 10-22
(e) the molecule of oxygen
Ans: 1 molecule of O2 has mass = $\frac{~~~~32~~}{6.023}\times $ 1023 = 5.314 $\times $ 10-23 g
(f) 0.25 gram atom of calcium
Ans: 0.25 gram atom of calcium has mass = 0.25 $\times $ 40 = 10g
13. Calculate the mass of 0.1 mole of each of the following
(Ca = 40, Na=23, Mg =24, S=32, C = 12, Cl = 35.5, O=16, H=1)
(a) CaCO3
Ans: 0.1 mole of CaCO3 has mass =100(molar mass) $\times $ 0.1 = 10 g
(b) Na2SO4.10H2O
Ans: 0.1 mole of Na2SO4.10H2O has mass = 322 $\times $ 0.1 = 32.2 g
(c) CaCl2
Ans: 0.1 mole of CaCl2 has mass = 111 $\times $ 0.1 = 11.1g
(d) Mg
Ans: 0.1 mole of Mg has mass = 24 $\times $ 0.1 = 2.4 g
14. Calculate the number of
(a) oxygen atoms in 0.10 mole of Na2CO3.10H2O.
Ans: 1molecule of Na2CO3.10H2O contains oxygen atoms = 13
So, 6.023 $\times $ 1023 molecules (1mole) has atoms = 13 $\times $ 6.023 $\times $ 1023
So, 0.1 mole will have atoms = 0.1 $\times $ 13 $\times $ 6.023 $\times $ 1023
=7.8 $\times $ 1023
(b) gram atoms in 4.6 gram of sodium
Ans: Given Na = 4.6 gm
Atomic mass= 23
No. of gram atoms of Na = $\frac{\text{ }\!\!~\!\!\text{ Mass}\,\text{of}\,\text{Na}}{\text{Atomic}\,\text{mass}\,\text{of}\,\text{Na}}$
= $\frac{4.6}{23}$ = 0.2
(c) moles in 12 g of oxygen gas
Ans: 32 g of oxygen gas = 1 mole
1 gram of oxygen gas = $\frac{~~1~~}{32}$ mole
Given that 12 g of oxygen gas
No: of moles = $\frac{given\,mas}{molar\,mass}$
= $\frac{~~~12~~~}{32}$ = 0.375 mole
15. What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?
Ans: 3.2 g of S has number of atoms = 6.023 $\times $1023$\times $$\frac{\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }3.2\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}{32}$
= 0.6023 $\times $ 1023
So, 0.6023 $\times $ 1023 atoms of Ca has mass=40 $\times $$\frac{\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }0.6023\times {{10}^{23}}}{6.023\times {{10}^{23}}}$
= 4g
16. Calculate the number of atoms in each of the following:
(a) 52 moles of He
Ans:No. of atoms = 52 $\times $6.023 $\times $1023 = 3.131 $\times $ 1025
(b) 52 amu of He
Ans: 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 52 g of He
Ans: 4 g of He has atoms = 6.023 $\times $1023
So, 52 g will have = 6.023 $\times $ 1023 $\times$$\frac{~~~52~~~}{4}$= 7.828 $\times $1024 atoms
17. Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.
Ans: Molecular mass of Na2CO3 = 106 g
106 g has 2 $\times $ 6.023 $\times $1023 atoms of Na
So, 5.3g will have = 2 $\times $ 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$= 6.022 $\times $1022 atoms
Number of atoms of C = 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$ = 3.01 $\times $ 1022 atoms
And atoms of O = 3 $\times $ 6.023 $\times $1023 $\times \frac{~~~5.3~~~}{106}$ = 9.03 $\times $ 1022 atoms
18. (a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea (CO(NH2)2) (O = 16; N = 14; C = 12 ; H = 1 )
Ans: 60 g urea has mass of nitrogen(N2) = 28 g
So, 5000 g urea will have mass = 28 $\times \frac{5000}{60}$ = 2.33 kg
(b) Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. (S = 32; O = 16)
Ans: 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 $\times $$\frac{320}{64}$ = 112 litres
19. (a) What do you understand by the statement that 'vapour density of carbon dioxide is 22'?
Ans: Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Atomic mass of Chlorine is 35.5.What is its vapour density?
Ans: Vapour density of Chlorine atom is 35.5.
20. What is the mass of 56 cm3 of carbon monoxide at STP?
(C=12, O=16)
Ans: 22400 cm3 of CO has mass = 28 g
So, 56 cm3 will have mass = 56 $\times \frac{28}{22400}$ = 0.07 g
21. Determine the number of molecules in a drop of water which weighs 0.09g.
Ans: 18 g of water has number of molecules = 6.023 $\times $ 1023
So, 0.09 g of water will have no. of molecules = 6.023 $\times $ 1023 $\times$ $\frac{0.09}{18}$ = 3.01 $\times $ 1021 molecules
22. The molecular formula for elemental sulphur is S8.In sample of 5.12 g of sulphur
(a) How many moles of sulphur are present?
Ans: No. of moles in 256 g S8 = 1 mole
So, no. of moles in 5.12 g = $\frac{~~5.12~~~}{256}$= 0.02 moles
(b) How many molecules and atoms are present?
Ans: No. of molecules = 0.02 $\times $ 6.023 $\times $ 1023 = 1.2 $\times $ 1022 molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in 1.2 $\times $ 1022 molecules = 1.2 x$\times $ 1022 $\times$ 8
= 9.635$\times $ 1022 molecules
23. If phosphorus is considered to contain P4 molecules, then calculate the number of moles in 100g of phosphorus?
Ans: Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P4 = 123.88 g
If phosphorus is considered as P4 molecules,
then 1 mole P4 ≡ 123.88 g
Therefore, 100 g of P4 = 0.807 g
24. Calculate:
(a) The gram molecular mass of chlorine if 308cm3 of it at STP weighs 0.979 g
Ans: 308 cm3 of chlorine weighs = 0.979 g
So, 22400 cm3 will weigh = gram molecular mass
= 0.979 $\times \frac{~~~22400~~}{308}$ =71.2 g
(b) The volume of 4g of H2 at 4 atmospheres.
Ans: 2 g(molar mass) H2 at 1 atm has volume = 22.4 litres
So, 4 g H2 at 1 atm will have volume = 44.8 litres
Now, at 1 atm(P1) 4 g H2 has volume (V1) = 44.8 litres
So, at 4 atm(P2) the volume(V2) will be = $\frac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}}$ = $\frac{1\times 44.8}{4}$ = 11.2 litres
(c) The mass of oxygen in 2.2 litres of CO2 at STP.
Ans: Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = 2.2 $\times \frac{~~~32~~}{22.4~}$=3.14 g
25. A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10-12 g, calculate the number of carbon atoms in the signature.
Ans: No. of atoms in 12 g C = 6.023 $\times $ 1023
So, no. of carbon atoms in 10-12 g = 10-12 $\times $$\frac{6.023\times{{10}^{23}}}{12}$
= 5.019 $\times $ 1010 atoms
26. An unknown gas shows a density of 3 g per litre at 2730C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?
Ans: Given:
P = 1140 mm Hg
Density = D = 2.4 g / L
T = 273 0C = 273+273 = 546 K
M = ?
We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.
First, apply Charle’s law.
We have to find out the volume of one litre of unknown gas at standard temperature 273 K.
V1= 1 L
T1 = 546 K
V2=?
T2 = 273 K
\[\frac{{{\text{V}}_{\text{1 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{1}}}}\]= $\frac{{{\text{V}}_{\text{2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }}}}{{{\text{T}}_{\text{2}}}}$
V2 = $\frac{{{V}_{1}}\times {{T}_{2}}}{{{T}_{1}}}$
= $\frac{1l\times 273}{546}$ = 0.5 L
We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.
Apply Boyle’s law.
P 1 = 1140 mm Hg
V1 = 0.5 L
P2 = 760 mm Hg
V2 = ?
P1 $\times $ V1 = P2 $\times $ V2
V2 = \[\frac{{{P}_{1}}\times {{V}_{1}}}{{{P}_{2}}}\]
=$\frac{1140\,mm\,hg\,\times \,0.5~L}{760mm\,hg}$ = 0.75 L
Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = \[\frac{0.75}{22.4}\]
= 0.0335 moles
The original mass is 2.4 g
n = $\frac{~~m~}{M}$ = 0.0335 moles
= 2.4 g / M
M = $\frac{~~~2.4~~~}{0.0335}$
M = 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g
27. Cost of Sugar (C12H22 O11) is Rs 40 per kg; calculate its cost per mole.
Ans: 1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost= $\frac{~~342\times 40~~~}{1000}$ =Rs. 13.68
28. Which of the following weighs the least?
(a) 2 g atom of N
(b) 3 x1025 atoms of carbon
(c) 1mole of sulphur
(d)7 g of silver
Ans: d) 7 g of silver
a. Weight of 1 g atom N = 14 g
So, weight of 2 g atom of N = 28 g
b. 6.023 $\times $1023 atoms of C weigh = 12 g
So, 3 $\times $1025 atoms will weigh = $\frac{~~~12\times 3\times {{10}^{25}}}{6.022~\times ~{{10}^{23}}}$=597.7g
c. 1 mole of sulphur weighs = 32 g
d. 7 g of silver
So, 7 grams of silver weighs the least.
29. Four grams of caustic soda contains:
(a) 6.02 x 1023 atoms of it
(b) 4 g atom of sodium
(c) 6.02 x1022 molecules
(d) 4 moles of NaOH
Ans: (c) 6.02 $\times $1022 molecule
40 g of NaOH contains 6.023 $\times $1023 molecules
So, 4 g of NaOH contains = $\frac{~~~6.02\times {{10}^{23}}\times 4~~~~}{40}$6.02 $\times $1023 $\times $ \[\frac{4}{40}\]
= 6.02 $\times $1022 molecules
30. The number of molecules in 4.25 g of ammonia is:
(a) 1.0 $\times $ 1023
(b) 1.5 $\times $ 1023
(c) 2.0 $\times $ 1023
(d) 3.5 $\times $ 1023
Ans: (b) 1.5 $\times $ 1023
The number of molecules in 18 g of ammonia = 6.02 $\times $1023
So, no. of molecules in 4.25 g of ammonia = $\frac{~~~6.02\times {{10}^{23}}\times ~4.25~~~}{18}$
= 1.5 $\times $ 1023
31. Correct the statements, if required
(a) One mole of chlorine contains 6.023 $\times $ 1010 atoms of chlorine.
Ans: One mole of chlorine contains 6.023 $\times $ 1010 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.
Ans: 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-12.
Ans: 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-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.
Ans: Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.
Exercise – 5C
1. Give three kinds of information conveyed by the formula H2O.
Ans: Information conveyed by H2O
(1)That H2O contains 2 volumes of hydrogen and 1 volume of oxygen.
(2)That ratio by weight of hydrogen and oxygen is 1:8.
(3)The molecular weight of H2O is 18g.
2. Explain the terms empirical formula and molecular formula.
Ans: The empirical formula 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.
3. Give the empirical formula of:
(a) C6H6
Ans: CH
(b) C6H12O6
Ans: CH2O
(c) C2H2
Ans: CH
(d) CH3COOH
Ans: CH2O
4. Find the percentage of water of crystallisation in CuSO4.5H2O. (At. Mass Cu = 64, H = 1, O = 16, S = 32)
Ans: Relative molecular mass of CuSO4. 5H2O
=64+32+4$\times $16+5(2+16)
=160+90=250
250g of CuSO4.5H2O contains 90g of water of crystallisation
=$\frac{90}{250}\times $100= 36%
5. Calculate the percentage of phosphorus in
(a) Calcium hydrogen phosphate Ca(H2PO4)2
Ans: Molecular mass of Ca(H2PO4)2 = 234
So, % of P = 2 $\times $ 31 $\times $ \[\frac{100}{234}\] = 26.5%
(b) Calcium phosphate Ca3(PO4)2
Ans: Molecular mass of Ca3(PO4)2 = 310
% of P = 2 $\times $31 $\times $ \[\frac{100}{310}\] = 20%
6. Calculate the percent composition of Potassium chlorate KClO3.
Ans: Molecular mass of KClO3 = 122.5 g
% of K = \[\frac{39}{122.5}\] = 31.8%
% of Cl = \[\frac{35.5}{122.5}\] = 28.98%
% of O = 3 $\times $ \[\frac{16}{122.5}\] = 39.18%
7. Find the empirical formula of the compounds with the following percentage composition:
Pb = 62.5%, N = 8.5%, O = 29.0%
Ans:
Element | % | At. mass | Atomic ratio | Simple ratio |
Pb | 62.5% | 207 | $\frac{~~~62.5~~~}{207}$=0.3019 | $\frac{~~0.3019~}{0.3019}$=1 |
N | 8.5% | 15 | $\frac{~~~~8.5~~~}{15}$= 0.6071 | $\frac{~~~0.6071~~}{0.3019}$=2 |
O | 29.0 | 16 | $\frac{~~~~~29~~~}{16}$= 1.81 | $\frac{~~1.81~~}{0.3019}$=6 |
So, Pb(NO3)2 is the empirical formula.
8. Calculate the mass of iron in 10 kg of iron ore which contains 80% of pure ferric oxide.
Ans: In Fe2O3, Fe = 56 and O = 16
Molecular mass of Fe2O3 = 2 $\times $56 + 3 $\times $16 = 160 g
Iron present in 80% of Fe2O3 = $\frac{~~112~~}{160}\times $80 = 56g
So, mass of iron in 100 g of ore = 56 g
mass of Fe in 10000 g of ore = 56 $\times $ \[\frac{10000}{100}\]
= 5.6 kg
9. If the empirical formula of two compounds is CH and their Vapour densities are 13 to 39 respectively, find their molecular formula.
Ans: For acetylene, molecular mass = 2 $\times $V.D = 2 $\times $13 = 26 g
The empirical mass = 12(C) + 1(H) = 13 g
n =$\frac{\text{Molecular formula mass}}{\text{Empirical formula weight}}$ = $\frac{26}{13}$ = 2
Molecular formula of acetylene= 2 $\times $Empirical formula =C2H2
Similarly, for benzene molecular mass= 2 $\times $V.D = 2 $\times $ 39 = 78
n = $\frac{78}{13}$ = 6
So, the molecular formula = C6H6
10. Find the empirical formula of a compound containing 17.7% hydrogen and 82.3% nitrogen.
Ans: Element % At. mass Atomic ratio Simple ratio
Element | % | Atomic mass | Atomic ratio | Simple ratio |
H | 17.7 | 1 | $\frac{~~~17.7~~}{1}$= 17.7 | $\frac{~~17.7~~}{5.87}$=3 |
N | 82.3 | 14 | $\frac{~~82.3~~}{14}$= 5.87 | $\frac{~~~5.87~~}{5.87}$=1 |
So, the empirical formula = NH3
11. On analysis, a substance was found to contain
C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44, calculate;
(a) its empirical formula, and
(b) its molecular formula
Ans:
Element | % | Atomic mass | Atomic ratio | Simple ratio |
C | 54.54 | 2 | $\frac{~~54.54~~}{12}$=4.55 | $\frac{~~4.55~~}{2.27}$=2 |
H | 9.09 | 4 | $\frac{~~~9.09~~}{1}$=9.09 | $\frac{~~~~9.09~~}{2.27}$= 4 |
O | 36.36 | 1 | $\frac{~~36.36~~}{16}$= 2.27 | $\frac{~~~2.27~~}{2.27}$=1 |
(a) So, its empirical formula = C2H4O
(b) empirical formula mass = 44
Since, vapour density = 44
So, molecular mass = 2 $\times $V.D = 88
or n = 2
so, molecular formula = (C2H4O)2 = C4H8O2
12. An organic compound, whose vapour density is 45, has the following percentage composition
H=2.22%, O = 71.19% and remaining carbon.
Calculate,
(a) its empirical formula, and
(b) its molecular formula
Ans:
Elements | % | at. mass | atomic ratio | simple ratio |
C | 26.59 | 12 | $\frac{~~~26.59~~}{12}$=2.21 | $\frac{~~2.21~~}{2.21}$= 1 |
H | 2.22 | 1 | $\frac{~~~~2.22~~~}{1}$=2.22 | $\frac{~~~2.22~~}{2.21}$= 1 |
O | 71.19 | 16 | $\frac{~~~71.19~~}{16}$= 4.44 | $\frac{~~~4.44~~}{2}$= 2 |
(a) its empirical formula = CHO2
(b) empirical formula mass = 45
Vapour density = 45
So, molecular mass = V.D $\times $2 = 90
so, molecular formula = C2H2O4
13. An organic compound contains H = 4.07%, Cl = 71.65% chlorine and remaining carbon. Its molar mass = 98.96. Find,
(a) Empirical formula, and
(b) Molecular formula
Ans:
Element | % | Atomic mass | Atomic ratio | Simple ratio |
Cl | 71.65 | 35.5 | $\frac{~~71.65~~}{35.5}$=2.01 | $\frac{~~~2.01~~}{2.02}$=1 |
H | 4.07 | 1 | $\frac{~~~4.07~~}{1}$=4.07 | $\frac{~~~4.07~~}{2.01}$=2 |
C | 24.28 | 12 | $\frac{~~~24.28~~~}{12}$=2.02 | $\frac{~~~24.28~~}{12}$=1 |
(a) its empirical formula = CH2Cl
(b) empirical formula mass = 49.5
Since, molecular mass = 98.96
so, molecular formula = (CH2Cl)2 = C2H4Cl2
14. A hydrocarbon contains 4.8g of carbon per gram of hydrogen. Calculate
(a) the g atom of each
Ans: The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1=1
(b) find the empirical formula
Ans:
Element | Given mass | At. mass | Atomic ratio | Simplest ratio |
C | 4.8 | 12 | $\frac{~~~4.8~~}{12}$=0.4 | $\frac{~~~0.4~~}{0.4~}$=1 |
H | 1 | 1 | $\frac{~~1~~}{1}$=1 | $\frac{~~1~~}{0.4}$=$\frac{~~5~~}{2}$ |
So, the empirical formula = C2H5
(c) Find molecular formula, if its vapour density is 29.
Ans: Empirical formula mass = 29
Molecular mass = V.D $\times $2 = 29 $\times $ 2 = 58
So, molecular formula = C4H10
15. 0.2 g atom of silicon Combine with 21.3 g of chlorine. Find the empirical formula of the compound formed.
Ans: 0.2g atom i.e 0.2 mole silicon
Moles of chlorine = $\frac{given\,mass}{molar\,mass}$ = $\frac{21.3}{35.5}$ = 0.6mol
Ratio of Si : Cl = $\frac{0.2}{0.2}$ : $\frac{0.6}{0.2}$ = 1:3
Empirical formula = SiCl3
16. A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.
Ans: % of carbon = 82.76%
% of hydrogen = 100 - 82.76 = 17.24%
Element | % Weight | Atomic Weight | Relative No. of Moles | Simplest Ratio |
C | 82.76 | 12 | $\frac{~~~82.76~~}{12}$= 6.89 | $\frac{~~~6.89~~~}{6.89}$= 1 $\times $ 2 = 2 |
H | 17.24 | 1 | $\frac{~~~17.24~~~}{17.24}$=1 | $\frac{~~~17.24~~~}{6.89}$= 2.5 $\times $ 2 = 5. |
Empirical formula = C2H5
Empirical formula weight = 2 $\times $ 12 + 1 $\times $ 5 = 24 + 5 = 29
Vapour Density = 29
Relative molecular mass = 29 $\times $ 2 = 58
N = $\frac{\text{Relative molecular mass}}{\text{Empirical}\,\text{weight}}$ = $\frac{58}{29}$ = 2
Molecular Formula = n x empirical formula
= 2 $\times $ C2H5
= C4H10
17. In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen. Deduce the simplest formula by answering the following questions.
(a) How many gram- atoms of magnesium are equal to 18g?
Ans: G atoms of magnesium = $\frac{18}{24}$ = 0.75 or g- atom of Mg
(b) How many gram- atoms of nitrogen are equal to 7g of nitrogen?
Ans: G atoms of nitrogen = $\frac{7}{14}$ = 0.5 or 1/2 g- atoms of N
(c) Calculate simple ratio of gram- atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.
Ans: Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg3 N2
18. Barium chloride crystals contain 14.8% water of crystallization. Find the number of molecules of water of crystallization per molecule.
Ans: Barium chloride = BaCl2. x H2O
Ba + 2Cl + x(H2 + O)
=137+ 235.5 + x(2+16)
= (208 + 18x) contains water = 14.8% water in BaCl2.x H2O
= (208 + 18 x) $\times \frac{14.8}{100}$ = 18x
= (104 + 9x) 2148=18000x
= (104+9x) 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2 molecules of water
19. Urea is a very important nitrogenous fertilizer. Its formula is CON2H4.Calculate the percentage of nitrogen in urea. (C=12, O=16, N=14 and H=1).
Ans: Molar mass of urea; CON2H4 = 60 g
So, % of Nitrogen = 28 $\times \frac{100}{60}$ = 46.66%
20. Determine the formula of the organic compound if its molecule contains 12 atoms of carbon. The percentage compositions of hydrogen and oxygen are 6.48 and 51.42 respectively.
Ans:
Element | % | Atomic mass | Atomic ratio | Simple ratio |
C | 42.1 | 12 | $\frac{~~~42.1~~}{12}$=3.5 | $\frac{~~3.5~~}{3.2}$=1 |
H | 6.48 | 1 | $\frac{~~~~6.48~~}{1}$=6.48 | $\frac{~~~6.48~~}{3.2}$=2 |
O | 51.42 | 16 | $\frac{~~~51.42~~}{16}$=3.2 | $\frac{~~3.2~~}{3.2}$=1 |
The empirical formula is CH2O
Since the compound has 12 atoms of carbon, so the formula is
C12 H24 O12.
21. (a) A compound with empirical formula AB2, has the vapour density equal to its empirical formula weight. Find its molecular formula.
Ans: Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, the molecular formula is A2B4.
(b) A compound with empirical formula AB has vapour density 3 times its empirical formula weight. Find the molecular formula.
Ans: Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = $\frac{\text{V}.\text{D}}{3}$
So, n = $\frac{\text{Molecular mass}}{\text{Empirical formula}\,\text{weight}}$ = 6
Hence, the molecular formula is A6B6
(c)10.47 g of a compound contains 6.25 g of metal A and rest non-metal B. Calculate the empirical formula of the compound (At. wt of A = 207, B = 35.5)
Ans: Wt. of the compound: 10.47g
Wt. of metal A: 6.25g
Wt. of non-metal B: 10.47 – 6.25 = 4.22g
Element | mass | At. Wt. | Relative no. of atoms | Simplest ratio |
A | 6.25g | 207 | $\frac{~~~6.25~~}{207}$= 0.03 | $\frac{~~~0.03~~}{0.03}$=1 |
B | 4.22g | 35.5 | $\frac{~~~4.26~~}{35.5}$=0.12 | $\frac{~~~0.12~~}{0.03}$=4 |
Empirical formula AB4
22. A hydride of nitrogen contains 87.5% percent by mass of nitrogen. Determine the empirical formula of this compound.
Ans: Atomic ratio of N = $\frac{87.5}{14}$ = 6.25
Atomic ratio of H = $\frac{12.5}{1}$ = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH2
23. A compound has O=61.32%, S= 11.15%, H=4.88% and Zn=22.65%.The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallization.
Ans:
Element | % | Atomic mass | Atomic ratio | Simple ratio |
Zn | 22.65 | 65 | $\frac{~~~~22.65~~}{65}$=0.348 | $\frac{~~0.348~~}{0.348}$=1 |
H | 4.88 | 1 | $\frac{~~4.88~~~}{1}$=4.88 | $\frac{~~4.88~~}{0.348}$=14 |
S | 11.15 | 32 | $\frac{~~~11.15~~}{32}$=0.348 | $\frac{~~~0.348~~}{0.348}$=1 |
O | 61.32 | 16 | $\frac{~~61.32~~}{16}$=3.83 | $\frac{~~~~3.83~~}{0.348}$=11 |
Empirical formula of the given compound =ZnSH14O11
Empirical formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
\[ n = \frac{\text{Molecular mass}}{\text{Empirical Formula}} = \frac{287}{287} = 1\]
Molecular formula = ZnSO11H14
=ZnSO4.7H2O
Exercise – 5D
1. Complete the following blanks in the equation as indicated.
CaH2 (s) + 2H2O (aq) $\to $Ca(OH)2 (s) + 2H2 (g)
(a) Moles: 1 mol + ------- $\to $ -------- + --------------
Ans: (a) Moles:1 mole + 2 mole \[\rightarrow\]1 mole + 2 mole
(b) Grams: 42g + ------- $\to $ -------- + ----------------
Ans: Grams: 42g + 36g \[\rightarrow\] 74g + 4 g
(c) Molecules: 6.02 x 1023 + ------- $\to $-------- + -----------
Ans: Molecules = 6.02 \[\times\] 1023 + 12.046 \[\times\] 1023 \[\rightarrow\] 6.02 \[\times\] 1023+ 12.046 \[\times\] 1023
2. The reaction between 15 g of marble and nitric acid is given by the following equation:
CaCO3 + 2HNO3$\to $Ca(NO3)2+ H2O + CO2
Calculate:
(a) the mass of anhydrous calcium nitrate formed
Ans: 100 g of CaCO3 produces = 164 g of Ca(NO3)2
So, 15 g CaCO3 will produce = 164 \[\times\] 15/100 = 24.6 g Ca(NO3)2
(b) the volume of carbon dioxide evolved at S.T.P.
Ans: 1 V of CaCO3 produces 1 V of CO2
100 g of CaCO3 has volume = 22.4 litres
So, 15 g will have volume = 22.4 \[\times\] 15/100 = 3.36 litres CO2
3. 66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.
Write a balanced equation and calculate:
2NH3 + H2SO4\[\rightarrow\](NH4)2SO4
(a) Mass of ammonia required.
Ans: 2NH3 + H2SO4 → (NH4)2SO4
34g 98g 132g
For 132 g (NH4)2SO4 = 34 g of NH3 is required
So, for 66 g (NH4)2SO4 = 66 × 32/132 = 17 g of NH3 is required
(b) The volume of the gas used at the S.T.P.
Ans: 17g of NH3 requires volume = 22.4 litres
(c) The mass of acid required.
Ans: Mass of acid required, for producing 132g (NH4)2SO4 = 98g
So, Mass of acid required, for 66g (NH4)2SO4 = 66 × 98/132 = 49g
4. The reaction between red lead and hydrochloric acid is given below:
Pb3O4 + 8HCl $\to $ 3PbCl2 + 4H2O + Cl2
Calculate
(a) the mass of lead chloride formed by the action of the 6.85 g of red lead,
Ans: Molecular mass of Pb3O4 = 3 × 207.2 + 4 × 16 = 685 g
685 g of Pb3O4 gives = 834 g of PbCl2
Hence, 6.85 g of Pb3O4 will give = 6.85 × 834/685 = 8.34 g
(b) the mass of the chlorine
Ans: 685g of Pb3O4 gives = 71g of Cl2
Hence, 6.85 g of Pb3O4 will give = 6.85 × \[\frac{71}{685}\] = 0.71 g Cl2
(c) the volume of the chlorine evolved at S.T.P.
Ans: 1 V Pb3O4 produces 1 V Cl2
685g of Pb3O4has volume = 22.4 litres = volume of Cl2 produced
So, 6.85 Pb3O4 will produce = 6.85 × \[\frac{22.4}{685}\] = 0.224 litres of Cl2
5. Find the mass of KNO3 required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO3 is required for the same purpose.
KNO3+ H2So4 $\to$ KHSO4 + HNO3
NaNO3 + H2SO4 $\to$ NaHSO4 + HNO3
Ans: Molecular mass of KNO3 = 101 g
63 g of HNO3 is formed by = 101 g of KNO3
So, 126000 g of HNO3 is formed by = 126000 × \[\frac{101}{63}\] = 202 kg
Similarly,126 g of HNO3 is formed by 170 kg of NaNO3
So, a smaller mass of NaNO3 is required.
6. Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27oC and normal pressure.
CaCO3 + 2HCl →CaCl2 + H2O + CO2
Calculate:
(a) The mass of salt required.
Ans: CaCO3 + 2HCl → CaCl2 + H2O + CO2
100g 73g 22.4L
According to given reaction
2L of O2 = $\frac{~~~2~~}{22.4}$ of CaCO3
Mass of salt required= 100$\times \frac{~~~2~~}{22.4~}$ = 8.93g
(b) The mass of the acid required to prepare the 2 litres of CO2 at 27 C and normal pressure.
CaCO3 + 2HCl $\to $ CaCl2 + H2O + CO2
Ans: 2L CO2 given by = 2$\times \frac{~~2~~}{22.4~}$mole HCl
Mass of HCl = 36.5$\times \frac{~~~4~~}{22.4}$ = 6.52g
7. Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water
Ans: 2H2O $\to $ 2H2 + O2
2 moles of H2O gives = 1 mole of O2
So, 1 mole of H2O will give = 0.5 moles of O2
so, mass of O2 = no. of moles x molecular mass
= 0.5 $\times $32 = 16 g of O2
and 1 mole of O2 occupies volume =22.4 litre
so, 0.5 moles will occupy = 22.4 × 0.5 = 11.2 litres at S.T.P.
8. 1.56 g of sodium peroxide reacts with water according to the following equation:
2Na2O2 + 2H2O$\to $4NaOH + O2
Calculate:
(a) mass of sodium hydroxide formed,
Ans: Mol. Mass of Na2O2 = 2 $\times $23 + 2 $\times $16 = 78 g
Mass of 2Na2O2= 156 g
156 g Na2O2 gives = 160 g of NaOH (4 $\times $40 g)
So, 1.56 Na2O2 will give = 160 $\times $ \[\frac{1.56}{156}\] = 1.6 g
(b) Volume of oxygen liberated at S.T.P.
Ans: 156 g Na2O2 gives = 22.4 litres of oxygen
So, 1.56 g will give = 22.4 $\times $1.56/156 = 0.224 litres
= 224 cm3
(c) Mass of oxygen liberated.
Ans: 156 g Na2O2 gives = 32 g O2
So, 1.56 g Na2O2 will give = 32 $\times $1.56/156
= 32/100 = 0.32 g
9. (a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH4Cl by the reaction:
2NH4Cl + Ca(OH)2 $\to$ CaCl2 +2H2O + 2NH3
Ans: 2NH4Cl + Ca(OH)2 $\to$ CaCl2+2H2O + 2NH3
Mol. Mass of 2NH4Cl = 2(14 + (1 $\times $ 4) + 35.5)= 2(53.5) = 107 g
107 g NH4Cl gives = 34 g NH3
So, 21.4 g NH4Cl will give = 21.4 $\times $34/107 = 6.8 g NH3
(b) What will be the volume of ammonia when measured at S.T.P?
The molar volume of a gas = 22.4 litres at STP.
Ans: The volume of 17 g NH3 is 22.4 litre
So, volume of 6.8 g will be = 6.8 $\times $22.4/17 = 8.96 litre
10. Aluminium carbide reacts with water according to the following equation.
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?
Ans: Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
1V 3V
144g 3$\times $22.4 Vol
Since 144g of Al4C3 gives 312g of Al(OH)3
So, 12g of Al4C3 will give = $\frac{312\times 12}{144}$ = 26g Al(OH)3
(b) What volume of methane s.t.p. is obtained from 12g of aluminium carbide?
Ans: 144g of Al4C3 gives 3$\times $22.4 litre of CH4
So 12g of Al4C3 will give = $\frac{3\times 22.4\times 12}{144}$ = 5.6 litre CH4
11. MnO2 + 4HCl $\to $MnCl2 + 2H2O +Cl2
0.02 moles of pure MnO2is heated strongly with conc. HCl. Calculate:
(a) mass of MnO2 used
Ans: 1 mole of MnO2 weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 $\times $ 0.02 = 1.74 g MnO2
(b) moles of salt formed,
Ans: 1 mole MnO2 gives = 1 mole of MnCl2
So, 0.02 mole MnO2will give = 0.02 mole of MnCl2
(c) mass of salt formed,
Ans: 1 mole MnCl2 weighs = 126 g(mol mass)
So, 0.02 mole MnCl2 will weigh = 126 $\times $ 0.02 g = 2.52 g
(d) moles of chlorine gas formed,
Ans: 0.02 mole MnO2will form =0.02 mol of Cl2
(e) mass of chlorine gas formed,
Ans: 1 mole of Cl2 weighs = 35.5 g
So, 0.02 mole will weigh = 71 $\times $ 0.02 = 1.42 g of Cl2
(f) volume of chlorine gas formed at S.T.P.,
Ans: 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = 22.4 $\times $ 0.02 = 0.448 litre
(g) moles of acid required,
Ans: 1 mole MnO2requires HCl = 4 mole
So, 0.02 mole MnO2 will require = 4 $\times $ 0.02 = 0.08 mole
(h) Mass of acid required.
Ans: For 1 mole MnO2 , acid required = 4 mole of HCl
So, for 0.02 mole, acid required = 4 $\times $ 0.02 = 0.08 mole
Mass of HCl = 0.08 $\times $ 36.5 = 2.92 g
12. Nitrogen and hydrogen react to form ammonia.
N2 (g) + 3H2 (g) $\to $ 2NH3 (g)
If 1000g H2 react with 2000g of N2:
(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?
Ans: N2 + 3H2 → 2NH3
28g 6g 34g
28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28 $\times $ 2000 = 3000/7g
So mass of hydrogen left unreacted = 1000 - \[\frac{3000}{7}\] = 571.4g of H2
(b) Calculate the mass of ammonia(NH3) that will be formed?
Ans: 28g of nitrogen forms NH3 = 34g
2000g of N2 forms NH3
= \[\frac{34}{28}\] $\times $ 2000
= 2428.6g
Miscellaneous Exercise
1. From the equation for burning of hydrogen and oxygen
2H2 + O2 $\to$ 2H2O (Steam)
Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.
Ans: As per the given equation
1 mole of Oxygen gives = 2 moles of steam
so, 0.5 mole oxygen will give = 2 $\times $0.5 = 1mole of steam
2. From the equation
3Cu + 8HNO3 → 3Cu (NO3)2+ 4H2O + 2NO
(At. mass Cu=64, H=1, N=14, O=16)
Calculate:
(a) Mass of copper needed to react with 63g of HNO3
Ans: Mol. Mass of 8HNO3 = 8 $\times $ 63 = 504 g
For 504 g HNO3, Cu required is = 192 g
So, for 63g HNO3Cu required = 192 $\times $ \[\frac{63}{504}\] = 24g
(b) Volume of nitric oxide at S.T.P. that can be collected.
Ans:504 g of HNO3 gives = 2 $\times $ 22.4 litre volume of NO
So, 63g of HNO3 gives =2 $\times $ 22.4 $\times $ \[\frac{63}{504}\] = 5.6 litre of NO
3. (a) Calculate the number of moles in 7g of nitrogen.
Ans: 28g of nitrogen = 1mole
So, 7g of nitrogen = \[\frac{1}{28}\] $\times $ 7 = 0.25 moles
(b) What is the volume at S.T.P. of 7.1 g of chlorine?
Ans: Volume of 71 g of Cl2 at STP = 22.4 litres
Volume of 7.1 g chlorine = 22.4 $\times $ \[\frac{7.1}{71}\] = 2.24 litre
(c) What is the mass of 56 cm3 of carbon monoxide at S.T.P?
Ans: 22400cm3 volume have mass = 28 g of CO(molar mass)
So, 56cm3 volume will have mass = 28 $\times $ \[\frac{56}{22400}\] = 0.07 g
4. Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?
Ans:
% of N in NaNO3 = $\frac{~~~14~~~}{85}\times $ 100= 16.47%
% of N in (NH4)2SO4 = $\frac{~~~~~14~~~}{132}\times $100= 21.21%
% of N in CO(NH2)2 =$\frac{~~~~14~~~}{60}$ $\times $ 100= 46.66%
So, the highest percentage of N is in urea.
5. Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:
2H2O$\to $2H2+O2
(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?
Ans: 2H2O $\to $ 2H2 + O2
2V 2V 1V
From equation, 2 V of water gives 2 V of H2 and 1 V of O2
where 2 V = 2500 cm3
so, volume of O2 liberated = \[\frac{\text{2V}}{\text{V}}\] = 1250 cm3
(b) The 2500 cm3 of H2 is subjected to 2 times increase in pressure (temp. remaining constant). What volume of H2 will now occupy?
Ans: $\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}{{{\text{T}}_{\text{1}}}}$=$\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$
$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$ = $\frac{\text{7}{{\text{P}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{\text{V}}_{\text{2}}}}{\text{2 }\!\!\times\!\!\text{ }{{\text{T}}_{\text{1}}}}$
V2= $\frac{~~~2500\times 2~~~}{7}$
V2= $\frac{~~5000~~}{7}$cm3
(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.
Ans:$\frac{~~{{V}_{1~~}}}{{{V}_{2}}}$ = $\frac{~~~~{{T}_{1~~~}}}{{{T}_{2}}}$
$\frac{~~~5000~~~}{7\times 2500}$= $\frac{~~~~{{\mathbf{T}}_{1~~~}}}{{{\mathbf{T}}_{2}}}$
T2= 3.5 T1
i.e. temperature should be increased by 3.5 times.
6. Urea (CO(NH2)2) is an important nitrogenous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?
Ans: Molecular mass of urea=12 + 16+2(14+2) =60g
60g of urea contains nitrogen =28g
So, in 50g of urea, nitrogen present =23.33 g
50 kg of urea contains nitrogen=23.33kg
7. Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.
Ans:
% of hydrogen = 20%
% of carbon = 100 - 20 = 80%
% Weight | Atomic Weight | Relative No. of Moles | Simplest Ratio | |
C | 80 | 12 | $\frac{~~~80~~}{12}$= 6.667 | $\frac{~~~6.667}{6.667}$= 1 |
H | 20 | 1 | $\frac{~~~20~~}{1}$= 20 | $\frac{~~~20~~}{6.667}$= 2.99 ≈ 3 |
Empirical formula = CH3
Empirical formula weight = 1 $\times $ 12 + 1 $\times $ 3 = 12 + 3 = 15
Vapour Density = 15
Relative molecular mass = 15 $\times $ 2 = 30
N = $\frac{\text{Relative molecular mass}}{\text{Empirical weight}}$= $\frac{30}{15}$ = 2
Molecular formula = n x empirical formula
= 2 $\times $ CH3
= C2H6
8. 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.
Ans: 22400cm3 CO2 has mass = 44g
so, 224 cm3 CO2 will have mass= 0.44 g
Now since CO2 is being formed and X is a hydrocarbon so it contains C and H.
In 0.44g CO2, mass of carbon=0.44-0.32=0.12g=0.01g atom
So, mass of Hydrogen in X = 0.145-0.12 = 0.025g
= 0.025g atom
Now the ratio of C:H is C=1: H=2.5 or C=2 : H=5
i.e. the formula of hydrocarbon is C2H5
(a) Which elements does X contain?
Ans: C and H
(b) What was the purpose of copper (II) oxide?
Ans: Copper (II) oxide was used for reduction of the hydrocarbon.
(c ) Calculate the empirical formula of X by the following steps:
(i) Calculate the number of moles of carbon dioxide gas.
Ans: no. of moles of CO2= 0.44/44 = 0.01 moles
(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.
Ans: mass of C = 0.12 g
(iii) Calculate the mass of hydrogen in sample X.
(iii) mass of H = 0.025 g
(iv) Deduce the ratio of atoms of each element in X (empirical formula).
(iv) The empirical formula of X = C2H5
9. A compound is formed by 24g of X and 64g of oxygen. If atomic mass of X=12 and O=16, calculate the simplest formula of compound.
Ans: Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element | % | Atomic mass | Atomic ratio | Simplest ratio |
X | 27.3 | 12 | $\frac{~~~27.3~~}{12}$=2.27 | $\frac{~~~~2.27~~}{2.27}$=1 |
O | 72.7 | 16 | $\frac{~~~72.2~~}{16}$=4.54 | $\frac{~~~~4.54~~}{2.27}$=2 |
So simplest formula = XO2
10. A gas cylinder filled with hydrogen holds 5g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:
(a) Vapour density of gas X.
Ans:
V.D = mass of gas at STP/mass of equal volume of H2 \[ = \frac{85}{5}\] = 17
(b) Molecular weight of gas X.
Ans: Molecular mass = 17(V.D) $\times $ 2= 34g
11. (a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :
CO2 + C $\to $2CO
What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?
Ans: CO2 + C \[\rightarrow\] 2CO
1V 1V 2V
12 g of C gives = 44.8 litre volume of CO
So, 3 g of C will give = 11.2 litre of CO
(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.
Ans: 2CO + O2 $\to$ 2CO2
2V 1V 2V
(i) 2 V CO requires oxygen = 1 V
so, 24 cm3 CO will require = 24/2 =12 cm3
(ii) 2 x 22400 cm3 CO gives = 2 $\times $ 22400 cm3 CO2
so, 24cm3 CO will give = 24 cm3 CO2
12. How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO2 evolved:
2Ca(NO3)2 $\to$ 2CaO+4NO2+O2
Ans: Molecular weight of 2Ca(NO3)2= 2[40+2(14+48)]
=328g
Molecular weight of CaO =2(40+16)
=112g
a. 328g of Ca(NO3)2 liberates 4 moles of NO2
Ans: 328g of Ca(NO3)2 liberates 4$\times $22.4L of NO2 , 82g will liberate
$\frac{~~~4\times 22.4~\times 82~~~}{328}$
=22.4dm3 of NO2
b. 328 g of calcium nitrate gives 112g of CaO
82 g will give $\frac{~~~112\times 82~~~}{328}$
=28 g of CaO
13.The equation for the burning of octane is:
2C8H18 + 25O2 $\to$ 16CO2 + 18H2O
2V 25V 16V 18V
(i) How many moles of carbon dioxide are produced when one mole of octane burns?
Ans: 2 moles of octane gives = 16 moles of CO2
so, 1 mole octane will give = 8 moles of CO2
(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?
Ans: 1 mole CO2 occupies volume = 22.4 litre
so, 8 moles will occupy volume = 8 $\times $22.4 = 179.2 litre
(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?
Ans: 1 mole CO2 has mass = 44 g
so, 16 moles will have mass = 44 $\times $16 = 704 g
(iv) What is the empirical formula of octane?
Ans: Empirical formula is C4H9.
14. Ordinary chlorine gas has two isotopes 3517Cl and 3717Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.
Ans: The relative atomic mass of Cl \[=\frac{\text{(35}\times \text{3+1}\times \text{37)}}{4}=35.5\] amu
15. Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.
Ans: Mass of silicon in the given compound =5.6g
Mass of the chlorine in the given compound=21.3g
Total mass of the compound=5.6g+21.3g=26.9g
% of silicon in the compound = 56/26.9 \[\times\] 100 = 20.82%
% of chlorine in the compound = 21.2/26.9 \[\times\] 100 = 79.18%
Element | % | Atomic mass | Atomic ratio | Simplest ratio |
Si | 20.82 | 28 | $\frac{~~~20.82~~}{28}$=0.74 | $\frac{~~~0.74~~}{0.74}$=1 |
Cl | 79.18 | 35 | $\frac{~~~79.18~~}{35.5~}$=2.23 | $\frac{~~2.23~~~}{0.74~~}$=3 |
So the empirical formula of the given compound =SiCl3
16. An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47 %; oxygen = 59.26 %. Find the empirical formula of the acid and its molecular formula, given that its relative molecular mass is 162.
Ans:
Element | % | Atomic mass | Atomic ratio | Simplest ratio |
P | 38.72 | 31 | $\frac{~~~38.72~~}{31}$=1.23 | $\frac{~~~1.23~~}{1.23}$=1 |
H | 2.47 | 1 | $\frac{~~~2.47~~}{1}$=2.47 | $\frac{~~2.47~~~}{1.23}$=2 |
O | 59.26 | 16 | $\frac{~~~59.26~~}{16}$=3.70 | $\frac{3.70~~}{1.23~~}$=3 |
So, empirical formula is PH2O3 or H2PO3
Empirical formula mass = 31+ 2 $\times $1 + 3 $\times $16 = 81
The molecular formula is = H4P2O6, because n = 162/81=2
17. a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27 oC and 700 mm pressure. The molecular mass of 'A' is 60.
Ans: V1 = 10 litres V2=?
T1= 27+ 273 = 300KT2=273K
P1=700 mmP2 = 760 mm
Using the gas equation
\[ \frac{P_{1}V_{1}}{T_{1}} \] = \[ \frac{P_{2}V_{2}}{T_{2}}\]
\[V_{2}\] = \[ \frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}} \] = \[ \frac{700 \times 10 \times 273}{300 \times 760}\]
Molecular weight A= 60
So weight of 22.4 litres of A at STP=60g
Weight of A at STP=$\frac{~~~~700\times 10\times 273~~~}{300\times 760}\times \frac{~~~60~}{22.4}$= 22.45g
b) A gas occupied 360 cm3 at 870C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find its relative molecular mass.
Ans:
V$\times \frac{~~~760~}{273}$= $\frac{~~~360\times 380~~~}{360}$
V=$\frac{~~~360\times ~380~\times 273~~~}{760\times }$=136.5cm3
136.5cm3 of gas weigh= 0.546
22400cm3 of gas weight= $\frac{~~0.546~~}{136.5}\times $22400= 89.6 amu
Relative molecular mass= 89.6 amu
18. A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.
(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?
Ans: Molecular mass of CO2 = 12+ 2$\times $16 = 44 g
So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22
V.D = $\frac{\text{mass of certain amount of C}{{\text{O}}_{\text{2}}}}{\text{mass of equal volume of hydrogen}}$ = $\frac{\text{m}}{1}$
22 = $\frac{\text{m}}{1}$
So, mass of CO2 = 22 kg
(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
Ans: According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder = X
19. Following questions refer to one mole of chlorine gas.
(a) What is the volume occupied by this gas at S.T.P.?
Ans: The volume occupied by 1 mole of chlorine = 22.4 litre
(b) What will happen to the volume of gas, if pressure is doubled?
Ans: Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) What volume will it occupy at 273oC?
Ans: \[\frac{{{\text{V}}_{\text{1}}}}{{{\text{V}}_{2}}}=\frac{{{\text{T}}_{\text{1}}}}{{{\text{T}}_{2}}}\]
\[\frac{\text{22}\text{.4}}{{{\text{V}}_{2}}}=\frac{273}{546}\]
V2 = 44.8 litres
(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?
Ans: Mass of 1 mole Cl2 gas =35.5 $\times $ 2 =71 g
20. (a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.
Ans: Total molar mass of hydrated CaSO4.xH2O = 136+18x
Since 21% is water of crystallization, so
$\frac{~~~18x~~~~}{136+18x}$= $\frac{~~~21~~}{100}$
So, x = 2 i.e. water of crystallization is 2.
(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.
Ans: For 18 g water, vol. of hydrogen needed = 22.4 litre
So, for 1.8 g, vol. of H2 needed = 1.8 $\times $ \[\frac{\text{22}\text{.4}}{18}\] = 2.24 litre
Now 2 vols. of water = 1 vol. of oxygen
1 vol. of water =1/2 vol. of O2 = \[\frac{\text{22}\text{.4}}{2}\] = 11.2 lit.
18 g of water = 11.2 lit. of O2
1.8 g of water = \[\frac{11.2}{18}=\frac{18}{10}\] = 1.12 lit.
(c) How much volume will be occupied by 2g of dry oxygen at 270C and 740 mm pressure?
Ans: 32g of dry oxygen at STP = 22400cc
2g will occupy = \[\frac{\text{224002}}{32}\] = 1400cc
P1 = 760 mm P2 = 740mm
V1 = 1400cc V2 =?
T1 = 273 K, T2 = 27 +73 = 300K
$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$= $\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$
${{\text{V}}_{2}}=\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}{{\text{T}}_{2}}}{{{\text{T}}_{1}}{{\text{P}}_{2}}}$ = $\frac{760\times 1400\times 300}{273\times 740}$ = 1580cc
$\frac{1580}{1000}$ = 1.58 litres
(d) What would be the mass of CO2 occupying a volume of 44 litres at 250 C and 750 mm pressure?
Ans: P1= 750 mm P2=760 mm
V1= 44lit. V2=?
T1= 298K T2=273K
$\frac{{{\text{P}}_{1}}{{\text{V}}_{1}}}{{{\text{T}}_{1}}}$= $\frac{{{\text{P}}_{2}}{{\text{V}}_{2}}}{{{\text{T}}_{2}}}$
${{\text{V}}_{\text{2}}}\text{=}\frac{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}{{\text{P}}_{\text{2}}}}$ = $\frac{750\times 44\times 273}{298\times 760}$ = 39.78litre
22.4 litre of CO2 at STP has mass = 44g
39.78 litre of CO2 at STP has mass = $\frac{\text{44 }\!\!\times\!\!\text{ 39}\text{.78}}{\text{22}\text{.4}}$ = 78.14g
(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) $\to $AgCl (s) + NaNO3
Calculate the percentage of NaCl in the mixture.