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.
We have completed the present discussion on Mole concept. In the next chapter, we will see Rate of chemical reactions.
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
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 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.
• 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.
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.
(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 10.25The 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.
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