Frank Textbook Solutions: Mole Concept And Stoichiometry

 Page 1


Chapter 5. Mole Concept And Stoichiometry
Solution 1: 
1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a 
simple ratio to one another, and also to the volume of the gaseous product, provided all 
the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, 
equal volumes of all gases contain the same number of molecules’.
Solution 2: 
Solution 4: 
Solution 3: 
When stating the volume of a gas, the pressure and temperature should also be given because 
the volume of a gas is highly susceptible to slight change in pressure and temperature of the 
gas.
PAGE NO : 103 
Page 2


Chapter 5. Mole Concept And Stoichiometry
Solution 1: 
1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a 
simple ratio to one another, and also to the volume of the gaseous product, provided all 
the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, 
equal volumes of all gases contain the same number of molecules’.
Solution 2: 
Solution 4: 
Solution 3: 
When stating the volume of a gas, the pressure and temperature should also be given because 
the volume of a gas is highly susceptible to slight change in pressure and temperature of the 
gas.
PAGE NO : 103 
Solution 5: 
Page 3


Chapter 5. Mole Concept And Stoichiometry
Solution 1: 
1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a 
simple ratio to one another, and also to the volume of the gaseous product, provided all 
the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, 
equal volumes of all gases contain the same number of molecules’.
Solution 2: 
Solution 4: 
Solution 3: 
When stating the volume of a gas, the pressure and temperature should also be given because 
the volume of a gas is highly susceptible to slight change in pressure and temperature of the 
gas.
PAGE NO : 103 
Solution 5: 
Solution 6: 
Solution 7:
1. Gram atom: “The quantity of the element which weighs equal to its gram atomic mass is 
called one gram atom of that element”. 
For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram 
atom of hydrogen.
2. Gram mole: “A sample of substance with its mass equal to its gram molecular mass is 
called one gram molecule of this substance or one gram mole”. 
For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 
32g.
Page 4


Chapter 5. Mole Concept And Stoichiometry
Solution 1: 
1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a 
simple ratio to one another, and also to the volume of the gaseous product, provided all 
the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, 
equal volumes of all gases contain the same number of molecules’.
Solution 2: 
Solution 4: 
Solution 3: 
When stating the volume of a gas, the pressure and temperature should also be given because 
the volume of a gas is highly susceptible to slight change in pressure and temperature of the 
gas.
PAGE NO : 103 
Solution 5: 
Solution 6: 
Solution 7:
1. Gram atom: “The quantity of the element which weighs equal to its gram atomic mass is 
called one gram atom of that element”. 
For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram 
atom of hydrogen.
2. Gram mole: “A sample of substance with its mass equal to its gram molecular mass is 
called one gram molecule of this substance or one gram mole”. 
For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 
32g.
Solution 8: 
Solution 9: 
Empirical formula:“Empirical formula of a compound is the formula which gives the number of 
atoms of different elements present in one molecule of the compound, in the simplest numerical 
ratio”. 
Molecular formula: “Molecular formula of a compound denotes the actual number of atoms of 
different elements present in one molecule of the compound”.
Solution 10:
1. The empirical formula of C
6
H
6
 is: CH
2. The empirical formula of C
6
H
12
O
6
 is: CH
2
O.
3. The empirical formula of C
2
H
2
 is: CH
4. The empirical formula of CH
3
COOH is: CH
2
O.
Solution 11: 
Three pieces of information conveyed by the formula H
2
O is that:
1. It shows that there are 2 hydrogen atoms and 1oxygen atoms present in H
2
O.
2. The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
3. It represents one molecule of compound water.
Solution 12: 
Page 5


Chapter 5. Mole Concept And Stoichiometry
Solution 1: 
1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a 
simple ratio to one another, and also to the volume of the gaseous product, provided all 
the volumes are measured at the same temperature and pressure’.
2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, 
equal volumes of all gases contain the same number of molecules’.
Solution 2: 
Solution 4: 
Solution 3: 
When stating the volume of a gas, the pressure and temperature should also be given because 
the volume of a gas is highly susceptible to slight change in pressure and temperature of the 
gas.
PAGE NO : 103 
Solution 5: 
Solution 6: 
Solution 7:
1. Gram atom: “The quantity of the element which weighs equal to its gram atomic mass is 
called one gram atom of that element”. 
For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram 
atom of hydrogen.
2. Gram mole: “A sample of substance with its mass equal to its gram molecular mass is 
called one gram molecule of this substance or one gram mole”. 
For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 
32g.
Solution 8: 
Solution 9: 
Empirical formula:“Empirical formula of a compound is the formula which gives the number of 
atoms of different elements present in one molecule of the compound, in the simplest numerical 
ratio”. 
Molecular formula: “Molecular formula of a compound denotes the actual number of atoms of 
different elements present in one molecule of the compound”.
Solution 10:
1. The empirical formula of C
6
H
6
 is: CH
2. The empirical formula of C
6
H
12
O
6
 is: CH
2
O.
3. The empirical formula of C
2
H
2
 is: CH
4. The empirical formula of CH
3
COOH is: CH
2
O.
Solution 11: 
Three pieces of information conveyed by the formula H
2
O is that:
1. It shows that there are 2 hydrogen atoms and 1oxygen atoms present in H
2
O.
2. The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
3. It represents one molecule of compound water.
Solution 12: 
Solution 13: 
Solution 14: 
Solution 15:
1. Na
2
SO
4
.10H
2
O.
2. C
6
H
12
O
6
.
PAGE NO : 104