Chapter 10.7 - Molarity of Solutions

In the previous section, we saw the details about STP. We also saw how the mole concept is related to balanced chemical equations. In this section we will see Molarity of solutions.

Number of Moles in solutions

• Several chemical substances like acids, alkalies etc., are used in the form of solutions. 
• We know that solutions are obtained when a solute is dissolved in a solvent. 
• So we will want to know 'how much quantity of a solute' is dissolved in 'how much quantity' of a solvent. 
• If we have that information, we can write the 'concentration' of that solution. 

Calculation of 'concentration' is achieved as follows:
1. Find the number (n) of moles of the solute. 
2. Find the total volume (V) of the solution (in litres).
3. Take the ratio ⁿ⁄_V. This is the number of moles present in one litre. 
4. This 'number of moles present in one litre' is taken as the 'concentration' of the solution. 
5. This method of expressing the concentration of the solutions is termed molarity.

---

So molarity is the number of moles in one litre of the solution.
• If one litre of the solution contains 1 mol of the solute, it is called a 1 molar solution.       
    ♦ In short form it is written as: A 1 M solution.
• If one litre of the solution contains 2 mol of the solute, it is called a 2 molar solution.       
    ♦ In short form it is written as: A 2 M solution.
• If one litre of the solution contains 0.5 mol of the solute, it is called a 0.5 molar solution.       
    ♦ In short form it is written as: A 0.5 M solution.
• Another situation where we will get a 0.5 M solution:
    ♦ Let the number (n) of moles in a solution be 1
    ♦ Let the volume (V) of the solution be 2 litres
    ♦ Then molarity = ⁿ⁄_V = ¹⁄₂ = 0.5
    ♦ It is a 0.5 M solution.

---

Let us now see the method to prepare a 1 M solution of NaCl:
1. Let us make a 1 M solution, which has a total volume of 1 L. So V = 1 L.
2. If V = 1 L and molarity also is to be 1, there must be 1 mole of the solute. (since molarity = ⁿ⁄_V )
• So we must take 1 mole of NaCl.
3. The GMM of NaCl is (23 + 35.5) = 58.5 grams. So we must take 58.5 grams of NaCl so that, 1 mole of NaCl molecules will be present.
4. Take this 58.5 grams and dissolve it by adding small quantities of water. Gradually bring up the volume to 1 L
5. In this way, the total volume of the solution as a whole will be 1 L, and exact 1 mole of NaCl will be present in it.

---

Given below are different possibilities for obtaining solutions of various molarity:
(i) n = 1 mol, V = 1 L
molarity = ⁿ⁄_V = ¹⁄₁ = 1 M
(ii) n = 2 mol, V = 2 L
molarity = ⁿ⁄_V = ²⁄₂ = 1 M

---

Now we will see some solved examples:

Solved example 10.25
The molecular mass of NaOH is 40. The molecular mass of NaCl is 58.5 g
(a) To prepare a 500 mL of 0.1 M NaOH solution, how much NaOH should be taken
(b) Explain the method of preparing a 0.5 M NaCl solution 
Solution:
(a) 1. We want a 0.1 M solution, whose total volume is 500 mL. 
2. So we can write:
• Molarity = 0.1 = ⁿ⁄_0.5 (since 500 mL = 0.5 L) ⇒ n = 0.1× 0.5 = 0.05
• So 0.05 moles of NaOH must be present in the final volume of 500 mL 
3. GMM of NaOH is 40. So one mole of NaOH has a mass of 40 g
4. So 0.05 mole of NaOH will have a mass of 0.05×40 = 2 grams 
5. Thus we get: Mass of NaOH required to prepare a 500 mL of 0.1 M NaOH solution = 2 grams
(b) 1. We want a 0.5 M solution. Let the total volume be 1 L. 
2. So we can write:
• Molarity = 0.5 = ⁿ⁄₁ ⇒ n = 1× 0.5 = 0.5
• So 0.5 moles of NaCl must be present in the final volume of 1 L 
3. GMM of NaCl is 58.5. So one mole of NaCl has a mass of 58.5 g
4. So 0.5 mole of NaOH will have a mass of 0.5×58.5 = 29.25 grams 
5. Thus we get: Mass of NaCl required to prepare a one L of NaCl solution = 29.25 grams 
6. Take this 29.25 grams and dissolve it by adding small quantities of water. Gradually bring up the volume to 1 L
7. In this way, the total volume of the solution as a whole will be 1 L and exact 0.5 mole of NaCl will be present in it.

Solved example 10.26
The gram molecular mass of glucose (C₆H₁₂O₆) is 180 g. Find the mass of glucose dissolved in 500 mL of 1 M solution of glucose
Solution:
1. We have a 1 M solution, whose total volume is 500 mL. 
2. So we can write:
• Molarity = 1 = ⁿ⁄_0.5 (since 500 mL = 0.5 L) ⇒ n = 1× 0.5 = 0.5
• So 0.05 moles of C₆H₁₂O₆ is present in the final volume of 500 mL 
3. GMM of C₆H₁₂O₆  is 180. So one mole of C₆H₁₂O₆ has a mass of 180 g
4. So 0.5 mole of C₆H₁₂O₆ will have a mass of 0.5×180 = 90 grams 
5. Thus we get: Mass of C₆H₁₂O₆ present in a 500 mL of 1 M C₆H₁₂O₆ solution = 90 grams  

Solved example 10.27
You are given 100 g of NaOH, 200 mL of water, beakers and weighing balance. How will you prepare a 1 molar (1 M) solution of NaOH by taking required quantities from these?
Solution:
1. We do not have 1 L. What we have is only 200 mL. This is one fifth of a litre.
2. One mole of NaOH is (23 + 16 + 1) = 40 g
3. To get 1 M solution, 40 g must be dissolved in 1 L 
4. That is., 40 g must be dissolved in 1000 mL
5. So, to get 1 M solution using 1 mL water, (⁴⁰⁄₁₀₀₀) = 0.04 g must be dissolved
6. So, to get 1 M solution using 200 mL water, 0.04 × 200 = 8 g must be dissolved
7. Thus, for a 1 M solution, dissolve this 8 grams of NaOH by adding small quantities of water, until the volume become 200 mL.   

Solved example 10.28
500 mL of a 1 M solution of common salt is taken.
(a) How many grams of common salt are dissolved in this 500 mL?
(b) If this 500 mL is dituted with water to a volume of 2 L, what will be the molarity of the new 2 L solution?
Solution:
1. We are given 500 mL of a solution of common salt (NaCl)
2. The molarity of this 500 mL is 1 
3. 500 mL is 0.5 L. So V = 0.5 L
4. So we can write: molarity = ⁿ⁄_V  = ⁿ⁄_0.5 = 1
5. Thus we get n = 1×0.5 = 0.5 
6. That means, 0.5 mole of NaCl is dissolved in the 500 mL
7. 1 mole of NaCl = 58.5 g
8. So 0.5 mole of NaCl = 0.5 × 58.5 = 29.25 g. This is the solution for part 1.
9. If the solution is diluted to a volume of 2 L, we can write:
n = 0.5 and V = 2 L (since number of moles does not change)
10. So molarity = ⁿ⁄_V = ^0.5⁄₂ = 0.25 M. This is the solution for part 2

---

We have completed the present discussion on Mole concept. In the next chapter, we will see Rate of chemical reactions. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved

Chapter 10.6 - Standard Temperature and Pressure - STP

In the previous section, we saw how to convert 'number of moles' into mass. We also saw how to calculate the mass of a single atom or molecule. In this section we will see the relation between 'mole' and 'Standard temperature and pressure' in the case of gases.

• In solids and liquids, the molecules are packed close to each other. The distance between any two molecules is very small. 
    ♦ So the volume of a solid or a liquid can be considered as the total volume of the molecules
• In gases, the molecules are not packed close to each other. The distance between any two molecules is very large when compared to the size of the molecules. 
    ♦ So the volume of a gas cannot be considered as the total volume of it's molecules.
• Also, the molecules of the gases are always in motion. So they occupy the entire space of it's container. 
    ♦ So volume of a gas can be considered as the volume of it's container. 

---

• Now consider any two gases: Gas A & Gas B.  
• Let the molecules of Gas A have a greater mass than those of Gas B. This is shown in the fig.10.7 below:

1. Let us take a sample from each of the two gases. Let both the samples contain the equal number of molecules. 
2. If the number of molecules are the same, naturally, we will think that the Gas A will have greater volume because, it has large size molecules. 
3. But that is not the case. Let us see the reason:
• The molecules with lesser mass will move with a larger speed than those with heavier mass. 
    ♦ So the lighter molecules are able to reach greater distances. 
    ♦ That means, they will be able to occupy a greater volume, even if they are small in size.
4. In fact, it is seen that, if any two gases gases have the same number of molecules, they will occupy the same volume.
5. But, to occupy the same volume, there is one condition: 
• The temperature and pressure experienced by the two gases should be the same. 
6. This is because:
• If any one gas is subjected to a greater temperature than the other, it's molecules will begin to move with a greater speed. So that gas will occupy a greater volume
    ♦ For example, if the temperature around an inflated balloon is increased, it will expand
• If any one gas is subjected to a greater pressure than the other, it's volume will decrease. 
    ♦ For example, if we apply pressure on an inflated balloon, it will contract. 
• Thus it is clear that, the gases under consideration must be experiencing the same temperature and pressure. Only then we will get the same volume for equal number of molecules
7. Consider the words: 'equal number of molecules'. 
• Why not take the 'mole' as the equal number? 
Let us analyse:
• We know that, a mole will contain N_A number of molecules. It will definitely be convenient to take it as the 'equal number of molecules'.
8. So take a mole each of Gas A and Gas B. 
• Let both the gases be subjected to the same temperature: 273 K
• Let both the gases be subjected to the same pressure: 1 atm
■ In such a condition, each of the gases A and B, will occupy a volume of 22.4 L
9. 273 K temperature and 1 atm pressure are known as standard temperature and pressure. It is abbreviated as STP

---

Note: 
Units of Temperature: 
• In day to day life, we use the celsius scale for measuring temperature. 
• For scientific and engineering purposes, another scale is also used. It is the 'kelvin scale'. It's symbol is 'K'. 
• They can be easily converted into each other
Units of Pressure: 
• We know that pressure is 'Force per unit area'. In day to day life, we use pascal as the unit for pressure. It is 'newton per square metre'. It's symbol is 'pa'. 
• For scientific and engineering purposes, another scale is also used. It is the 'atmosphere scale'. It's symbol is 'atm' 
• They can be easily converted into each other
Units of Volume
• In day to day life we use 'Litre' to measure volume. It's symbol is 'L'. 
• For scientific and engineering purposes, another unit is also used. It is the 'Cubic centimetre'. It's symbol is cm³
• They can be easily converted into each other using the relation: 1 L = 1000 cm³
We will learn more details about such units in higher classes

---

In the above discussion, we considered two gases A and B. In fact, it is valid for any number of gases. That is:

• Take samples of 'n' gases. Where 'n' can be any number greater than 1:  2, 3, 4, 5 so on...

    ♦ Each sample should be subjected to a temperature of 273 K

    ♦ Each sample should be subjected to a pressure of 1 atm

• In other words, each sample should be at STP

• Further, each sample should contain exactly N_A number of it's molecule

■ Then each of those n samples will occupy a volume of 22.4 L

---

• N_A (Avogadro number) of molecules of a gas means 'one mole of molecules' of that gas. 

So we can write:
■ Take one mole of molecules of any gas. If those molecules are subjected to 273 K and 1 atm, they will occupy a volume of 22.4 L
■ Conversely, take 22.4 L of any gas. If the temperature and pressure experienced by that volume is 273 K and 1 atm, then there will be one mole of molecules in that volume.
So we can modify the presentation that we saw earlier. The modified presentation is given in fig.10.8 below:

Let us see an example:
A gas occupies a volume of 224 L when it is subjected to 273 K temperature and 1 atm pressure. How many molecules are present in that volume? 
Solution:
1. Temperature is 273 K and pressure is 1 atm. So the gas is at STP
2. At STP, 22.4 L of any gas will contain 1 mole of it's molecules
3. So 1 L will contain (¹⁄_22.4) mole of it's molecules
4. Thus, no. of moles in 224 L = 224 × ¹⁄_22.4 = 10
5. No. of molecules in 10 moles = 10 × 6.022×10²³

Solved example 10.21
Calculate the number of molecules in each of the following two samples:
(a) 44.8 L of NH₃ at STP  (b) 67.2 L of N₂ at STP

Solution:
(a) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (¹⁄_22.4) mol
• So 44.8 L will contain [44.8 × (¹⁄_22.4)] = 2 mol
• 1 mol will contain 6.022×10²³ molecules
• So 2 mol will contain (2 × 6.022×10²³) molecules
(b) 1 mol of any gas will occupy 22.4 L at STP
• So 1 L at STP will contain (¹⁄_22.4) mol
• So 67.2 L will contain [67.2 × (¹⁄_22.4)] = 3 mol
• 1 mol will contain 6.022×10²³ molecules
• So 3 mol will contain (3 × 6.022×10²³) molecules

Solved example 10.22
Calculate the volume of 170 g ammonia at STP
Solution:
1. The GMM of ammonia is [(1×14)+(3×1)] = [14+3] = 17
2. So 1 mole of ammonia will have a mass of 17 g
3. So 1 g will contain (¹⁄₁₇) mole 
4. So 170 g will contain [170×(¹⁄₁₇) ] = 10 mole 
5. 1 mole of any gas will have a volume of 22.4 L at STP
So 170 g (10 moles) of ammonia will have a volume of 22.4×10 = 224 L

Mole concept – In balanced chemical equations

In this topic,we will learn how the mole concept is related to balanced chemical equations. We have already learned about balanced chemical equations. Details here. 
Consider the balanced chemical equation for the manufacture of ammonia:
N₂ + 3H₂ → 2NH₃ 
• One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia. 
• So if we take 1 molecule of nitrogen, it is compulsory to take 3 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 2 molecules of ammonia.
• If we take 2 molecules of nitrogen, it is compulsory to take 6 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 4 molecules of ammonia.
• If we take 15 molecules of nitrogen, it is compulsory to take 45 molecules of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get 30 molecules of ammonia.
So on . . . 
• Thus, if we take N_A molecules of nitrogen, it is compulsory to take 3N_A molecules of hydrogen. After the reaction, we will get 2N_A molecules of ammonia. 
■ But N_A molecules is 1 mole molecules. So we can write:
If we take 1 mole of nitrogen, it is compulsory to take 3 moles of hydrogen. After the reaction, we will get 2 moles of ammonia.

---

Once we write the quantities in terms of moles, we will be able to write the quantities in terms of grams:

• 1 mole of nitrogen molecules will have a mass of GMM of nitrogen, which is equal to 28 grams
• 1 mole of hydrogen molecules will have a mass of GMM of hydrogen, which is equal to 2 grams
    ♦ so 3 moles of hydrogen will have a mass of (3 × 2) = 6 grams
■ If we take 28 grams of nitrogen, it is compulsory to take 6 grams of hydrogen. Then only the reactants can be completely used up. After the reaction, we will get (28+6) = 34 grams of ammonia.
Check:
• 1 GMM of ammonia = [(1×14)+(3×1)] = [14+3] = 17
• The product is 2 moles of ammonia, which is 2 GMM of ammonia, which is equal to (2 × 17) = 34 grams of ammonia.

---

In terms of litres:

• 1 mole of any gas = 22.4 L at STP
• So, if we take 22.4 L of nitrogen gas at STP, it is compulsory to take (3×22.4) = 67.2 L of hydrogen at STP. When the reaction is complete, we will get (2×22.4) = 44.8 L of ammonia at STP.

---

The above results can be written in a tabular form as shown below:

	Nitrogen	Hydrogen	Ammonia
Mole	1	3	2
Mass	28 grams	6 grams	34 grams
Volume	22.4 L	67.2 L	44.8 L

Solved example 10.23
Calculate the amount of hydrogen required to react with 500 g of nitrogen during the manufacture of ammonia.
Solution:
1. The balanced equation is: N₂ + 3H₂ → 2NH₃.
2. So for one mole of nitrogen, three moles of hydrogen is required
• One mole of nitrogen is 28 grams.
• One mole of hydrogen is 2 grams
    ♦ Three moles of hydrogen is (3×2) = 6 grams  
3. So for 28 grams of nitrogen, 6 grams of hydrogen is required 
4. So for 1 gram of nitrogen, (⁶⁄₂₈) gram of hydrogen is required
5. Thus for 500 grams of nitrogen, amount of hydrogen required = 500 × (⁶⁄₂₈) = 107.14 grams

Solved example 10.24
How many grams of N₂ is required to obtain 1000 L of ammonia at STP?
Solution:
1. One mole of ammonia at STP will have a volume of 22.4 L
2. So 1 L of ammonia at STP will have (¹⁄_22.4) mole
3. So 1000 L of ammonia at STP will have 1000 × ¹⁄_22.4 = ¹⁰⁰⁰⁄_22.4 mole
4. The balanced equation for the manufacture of ammonia is: N₂ + 3H₂ → 2NH₃.
5. 2 moles of ammonia can be obtained from one mole of nitrogen
6. So one mole of ammonia can be obtained from ¹⁄₂ mole of nitrogen
7. So ¹⁰⁰⁰⁄_22.4 mole of ammonia can be obtained from ¹⁰⁰⁰⁄_22.4 × ¹⁄₂ = ⁵⁰⁰⁄_22.4 mole of nitrogen
8. One mole of nitrogen is 28 grams
9. So ⁵⁰⁰⁄_22.4 mole of nitrogen is (⁵⁰⁰⁄_22.4 × 28) grams

---

In the next section we will see Molarity of solutions. 

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved

Chapter 10.5 - Conversion of 'Number of Moles' into Mass

In the previous section, we saw how to convert mass into 'number of moles'. We also saw how to convert a given number of atoms/molecules into 'number of moles'. In this section we will see the reverse. That is., conversion of 'number of moles' into mass.

■ First we will see the calculation of mass of atoms from no. of moles:
• We know that one mole of 'atoms of a particular element' will have a mass of GAM of that atom
• So we can write:
Eq.10.2:
Total mass of a sample of atoms = No. of moles in that sample × GAM
■ Now we will see the calculation of mass of molecules from no. of moles:
• We know that one mole of 'molecules of a particular compound' will have a mass of GMM of that molecule
• So we can write:
Eq.10.3:
Total mass of a sample of molecules = No. of moles in that sample × GMM

---

Now we will see some solved examples:

Solved example 10.9

Given that GMM of water is 18 g. Find the number of molecules in 180 grams of water. Also find the total number of atoms in this 180 g of water

Solution:

1. One mole of 'molecules of a particular compound' will consist of N_A molecules of that compound

2. And it will have a mass of GMM of that molecule

3. So one GMM of water will have N_A molecules

4. No. of GMMs in 180 g = ¹⁸⁰⁄₁₈ =10

5. Thus total no. of molecules = 10 × N_A = 10 × 6.022×10²³.

6. Each of these molecules will have two hydrogen atoms and one oxygen atom

7. So total number of hydrogen atoms = 2 × 10 × 6.022×10²³

8. Total number of oxygen atoms = 1 × 10 × 6.022×10²³.

Solved example 10.10

What will be the mass (in grams) of 10,00,000 molecules of water
Solution:

1. One mole of 'molecules of a particular compound' will consist of N_A molecules of that compound

2. And it will have a mass of GMM of that molecule

3. So one GMM of water will have N_A molecules
4. One GMM of water = 18 grams. 
5. So mass of one mole of water = 18 grams. 
6. That is., mass of N_A molecules = 18 grams
That is., mass of 6.022×10²³ molecules = 18 grams
7. So mass of one molecule = ¹⁸⁄_{(6.022 × 10²³)} grams
8. Thus mass of 10,00,000 molecules 
= ¹⁸⁄_{(6.022 × 10²³)} × 1000000 
= ^180000000⁄_{(6.022 × 10²³)} grams

Solved example 10.11

How many molecules are present in 710 g of chlorine gas (Cl₂)? What will be the total number of atoms in it?

Solution:
1. One mole of Cl₂ will contain N_A molecules.
2. That is., one GMM of Cl₂ (= 2×35.5 = 71 grams) will contain N_A molecules 
3. No. of GMMs in 710 grams = ⁷¹⁰⁄₇₁ = 10
4. So no. of molecules = 10×N_A = 10×6.022×10²³
5. Each Cl₂ molecule will contain two atoms. So no. of atoms = 2×10×6.022×10²³

Solved example 10.12

How many molecules are present in 90 g of glucose? What will be the total number of atoms in it?

Solution:
1. One mole of glucose (C₆H₁₂O₆) will contain N_A molecules.
2. That is., one GMM of C₆H₁₂O₆ [=(6×12)+(12×1)+(6×16) =180 grams] will contain N_A molecules 
3. No. of GMMs in 90 grams = ⁹⁰⁄₁₈₀ = 0.5
4. So no. of molecules = 0.5×N_A = 0.5×6.022×10²³
5. Each C₆ H₁₂O₆ molecule will contain (6+12+6 = 24) atoms. 
So no. of atoms = 24×0.5×6.022×10²³

Solved example 10.13

Calculate the mass of each of the following in grams:

(i) One nitrogen atom (ii) One nitrogen molecule (iii) One sulphuric acid molecule

Given GAMs: H = 1 g, C = 12 g, S = 32 g, O = 16 g, Cl = 35.5 g, N = 14 g

Solution of part (i):

1. One mole of nitrogen atoms will contain N_A atoms.

2. That is., one GAM (= 14 g) of N will contain N_A atoms

3. So, mass of N_A atoms of nitrogen is 14 grams

4. Thus, mass of one atom of nitrogen = ¹⁴⁄_NA = ¹⁴⁄_{(6.022 × 10²³)} grams  

Solution of part (ii):

1. One mole of nitrogen molecules will contain N_A molecules.

2. That is., one GMM (= 28 g) of N will contain N_A molecules

3. So, mass of N_A molecules of nitrogen is 28 grams

4. Thus, mass of one molecule of nitrogen = ²⁸⁄_NA = ²⁸⁄_{(6.022 × 10²³)} grams  

Solution of part (iii):

1. One mole of sulphuric acid (H₂SO₄) molecules will contain N_A molecules.

2. That is., one GMM [=(2×1)+(1×32)+(4×16) =98 grams] of H₂SO₄ will contain N_A molecules

3. So, mass of N_A molecules of H₂SO₄ is 98 grams

4. Thus, mass of one molecule of H₂SO₄ = ⁹⁸⁄_NA = ⁹⁸⁄_{(6.022 × 10²³)} grams  

---

Now we will see some solved examples based on what we have learned so far in this chapter

Solved example 10.14

Suppose 20 molecules of hydrogen are allowed to react with 20 molecules of oxygen to produce water. 

(a) Which of the reactant molecules get consumed first?

(b) The molecules of which reactant will remain unreacted? How many?

Solution:

The balanced chemical equation for the reaction is:

2H₂ + O₂ → 2H₂O.

1. For every one oxygen molecule, there must be two hydrogen molecules. So if there are 20 oxygen molecules, there must be 40 hydrogen molecules. 

2. But we have only 20 hydrogen molecules. All of them will be used up by 10 oxygen molecules.

3. The remaining 10 oxygen molecules do not have any hydrogen molecules to react with. So we can write:

• The hydrogen molecules get consumed first

• Some molecules of oxygen will remain unreacted. Number of oxygen molecules which will remain unreacted is 10

Solved example 10.15

Which among the following is used as the basis of expressing atomic mass?

(Hydrogen, carbon-12, carbon-14, Oxygen-16)

Solution: The basis for expressing atomic mass is the 'mass of carbon-12 atom'. (Details here)

Solved example 10.16

The atomic mass of helium is 4 and that of oxygen is 16. How many grams of helium is required to get as many number of atoms as are present in 40 grams of oxygen?

Solution:

■ Consider 'x' grams of helium. We want to satisfy the following condition:

The number of helium atoms in x grams of helium 

= The number of oxygen atoms in 40 grams of oxygen

1. GAM of oxygen = 16 grams.

2. So 16 g of oxygen will contain N_A atoms of oxygen

So number of atoms in 1 g of oxygen = ^N_A⁄₁₆ 

3. Thus number of atoms in 40 g of oxygen = ^N_A⁄₁₆×40 = 2.5N_A

4. GAM of helium = 4 grams.

5. So 4 g of helium will contain N_A atoms of helium

So number of atoms in 1 g of helium = ^N_A⁄₄

6. Thus number of atoms in x g of oxygen = ^N_A⁄₄ × x = 0.25xN_A 

7. (3) and (6) must be equal. So we can write: 2.5N_A = 0.25xN_A.

⇒ 2.5 = 0.25x ⇒ x = ^2.5⁄_0.25 ⇒ x = 10 g

Solved example 10.17

Find the gram atomic masses in each of the following

(a) 100 g of helium  (b) 200 g of oxygen  (c) 70 g of nitrogen  (d) 1 g of calcium

(Atomic masses: He = 4, O = 16, N = 14, Ca = 40)

Solution:

(a) Atomic mass of helium = 4 u

So GAM of helium = 4 g

No. of GAMs in 100 g of helium = ¹⁰⁰⁄₄ = 25

(b) Atomic mass of oxygen = 16 u

So GAM of oxygen = 16 g

No. of GAMs in 100 g of oxygen = ²⁰⁰⁄₁₆ = 12.5

(c) Atomic mass of nitrogen = 14 u

So GAM of nitrogen = 14 g

No. of GAMs in 100 g of nitrogen = ⁷⁰⁄₁₄ = 5

(d) Atomic mass of calcium = 40 u

So GAM of calcium = 40 g

No. of GAMs in 1 g of calcium = ¹⁄₄₀ = 0.025

Solved example 10.18

Calculate the gram molecular mass / gram formula mass of the following:

(a) HNO₃  (b) CaCl₂  (c) Na₂SO₄  (d) NH₄NO₃

(Gram atomic masses: H = 1 g,  N = 14 g,  O = 16 g,  Na = 23 g,  S = 32 g,  Cl = 35.5 g, Ca = 40 g)

Solution:

(a) GAM of HNO₃ = (1×1) + (1×14) + (3×16) = 1 + 14 + 48 = 63 g   

(b) GAM of CaCl₂ = (1×40) + (2×35.5) = 40 + 71 = 111 g   

(c) GAM of Na₂SO₄ = (2×23) + (1×32) + (4×16) = 46 + 32 + 64 = 142 g   

(d) GAM of NH₄NO₃ = (1×14) + (4×1) + (1×14) + (3×16) = 14 + 4 + 14 + 48 = 80 g   

Solved example 10.19

Given below are few samples:

(a) 400 g of water (H₂O)  (b) 400 g of carbon (C)  (c) 400 g og helium (He) (d) 400 g of hydrogen (H₂)  (e) 400 g of glucose (C₆ H₁₂O₆)

(i) Find the number of moles in each

(ii) Arrange the samples in the increasing order of their number of moles

(Gram molecular masses: He = 4 g,  C = 12 g,  H₂ = 2 g,  H₂O = 18 g,  C₆ H₁₂O₆ = 180 g)

Solution:

(a) One GMM of H₂O = 18 g

• So one mole of H₂O has a mass of 18 g. 

• In other words, if we take 18 grams of H₂O molecules, it will contain 1 mole

• So, if we take 1 g of H₂O molecules, it will contain ¹⁄₁₈ mole   

• Thus, no. of moles in 400 g = 400 × ¹⁄₁₈ = ⁴⁰⁰⁄₁₈ moles

(b) One GMM of C = 12 g

• So one mole of C has a mass of 12 g. 

• In other words, if we take 12 grams of C molecules, it will contain 1 mole

• So, if we take 1 g of C molecules, it will contain ¹⁄₁₂ mole   

• Thus, no. of moles in 400 g = 400 × ¹⁄₁₂ = ⁴⁰⁰⁄₁₂ moles

(c) One GMM of He = 4 g

• So one mole of He has a mass of 4 g. 

• In other words, if we take 4 grams of He molecules, it will contain 1 mole

• So, if we take 1 g of He molecules, it will contain ¹⁄₄ mole   

• Thus, no. of moles in 400 g = 400 × ¹⁄₄ = ⁴⁰⁰⁄₄ moles

(d) One GMM of H₂ = 2 g

• So one mole of H₂ has a mass of 2 g. 

• In other words, if we take 2 grams of H₂ molecules, it will contain 1 mole

• So, if we take 1 g of H₂ molecules, it will contain ¹⁄₂ mole   

• Thus, no. of moles in 400 g = 400 × ¹⁄₂ = ⁴⁰⁰⁄₂ moles

(e) One GMM of C₆H₁₂O₆ = 180 g

• So one mole of C₆H₁₂O₆ has a mass of 180 g. 

• In other words, if we take 180 grams of C₆H₁₂O₆ molecules, it will contain 1 mole

• So, if we take 1 g of C₆H₁₂O₆ molecules, it will contain ¹⁄₁₈₀ mole   

• Thus, no. of moles in 400 g = 400 × ¹⁄₁₈₀ = ⁴⁰⁰⁄₁₈₀ moles

(ii) Let us write the number of moles in order:

• Water - ⁴⁰⁰⁄₁₈,  • Carbon - ⁴⁰⁰⁄₁₂,  • Helium - ⁴⁰⁰⁄₄,  • Hydrogen - ⁴⁰⁰⁄₂,  • Glucose - ⁴⁰⁰⁄₁₈₀

• We find that, the all are fractions with the same numerator. 

• So, if we want an increasing order of number of moles, we must have a decreasing order of denominators. Thus we can write:

• Glucose - ⁴⁰⁰⁄₁₈₀,  • Water - ⁴⁰⁰⁄₁₈,  • Carbon - ⁴⁰⁰⁄₁₂,  • Helium - ⁴⁰⁰⁄₄,  • Hydrogen - ⁴⁰⁰⁄₂,

Solved example 10.20

Calculate the following:
(a) Number of moles in 1 kg of water
(b) Number of moles in 500 g of CaCO₃
(c) Number of moles and total number of atoms in 88 g of CO₂
(GMM/GFM: H₂O = 18 g,  CaCO₃ = 100 g,  CO₂ = 44 g, )
Solution:
(a) One GMM of H₂O = 18 g
• So one mole of H₂O has a mass of 18 g. 
• In other words, if we take 18 grams of H₂O molecules, it will contain 1 mole
• So, if we take 1 g of H₂O molecules, it will contain ¹⁄₁₈ mole   
• Thus, no. of moles in 1000 g = 1000 × ¹⁄₁₈ = ¹⁰⁰⁰⁄₁₈ moles
(b) One GMM of CaCO₃ = 100 g
• So one mole of CaCO₃ has a mass of 100 g. 
• In other words, if we take 100 grams of CaCO₃ molecules, it will contain 1 mole
• So, if we take 1 g of CaCO₃ molecules, it will contain ¹⁄₁₀₀ mole   
• Thus, no. of moles in 500 g = 500 × ¹⁄₁₀₀ = ⁵⁰⁰⁄₁₀₀ = 5 moles
(c) One GMM of CO₂ = 44 g
• So one mole of CO₂ has a mass of 44 g. 
• In other words, if we take 44 grams of CO₂ molecules, it will contain 1 mole
• So, if we take 1 g of CO₂ molecules, it will contain ¹⁄₄₄ mole   
• Thus, no. of moles in 88 g = 88 × ¹⁄₄₄ = ⁸⁸⁄₄₄ = 2 moles
• So no. of molecules = 2 × 6.022×10²³
• Each of these molecules have 3 atoms. (1 atom of carbon and 2 atoms of oxygen)
• So no. of atoms = 3 × 2 × 6.022×10²³

---

In the next section we will learn about STP

PREVIOUS      CONTENTS       NEXT

                        Copyright©2017 High school Chemistry lessons. blogspot.in - All Rights Reserved