In the previous section, we saw the basics of stoichiometry. In this section we will see reactions in solutions
• On many occasions, reactions are carried out in solutions
• For example, in a reaction, a 'solution of NaOH in water' is one of the reactant
• In such cases, we use any one of the following four methods:
A. Mass percent or weight percent
B. Mole fraction
C. Molarity
D. Molality
We will see each of them in detail:
A. Mass percent
(i) In this method, we first write the masses:
• Mass of the solute (m1)
• Mass of the solvent (m2)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total mass (m1+m2) of the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)^{th}}$ of the total mass is contributed by the solute
• This fraction when expressed as a percentage, gives the mass percent
• So we can write:
Eq.1.3: Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$
An example:
5 g of a substance A is added to 20 grams of a solvent B. What is the mass percent of A?
Solution:
• We have:
Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$
♦ Let mass of the substance A be m1
♦ Let mass of the solvent B be m2
• Substituting the values, we get:
Mass percent = $\mathbf\small{\left(\frac{5}{5+20}\right)\times 100=\left(\frac{1}{5}\right)\times 100=20\text{%}}$
B. Mole fraction
(i) In this method, we first write the 'number of moles':
• No. of moles of the solute (n1)
• No. of moles of the solvent (n2)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total number of moles (n1+n2)' in the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)^{th}}$ of the 'total number of moles' is contributed by the solute
(iv) This fraction is called the mole fraction
• Note that, we do not need to convert this fraction into percentage form
• So we can write:
Eq.1.4: Mole fraction = $\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)}$
C. Molarity
• Detailed notes on molarity can be seen here.
• Based on those notes, we can now derive some interesting results:
• Consider the following situation:
(i) A solution of NaOH is available in a lab
(ii) This solution is named as: Solution A
(iii) The molarity of A is: 1 M
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution of NaOH
(vi) This new solution is named as: Solution B
(vii) The molarity of B should be 0.2 M
(viii) How much is 'V'?
• The above 8 points describe the situation
• But two points seem to be missing:
♦ What is the total volume of A?
♦ What is the total volume of B?
• For calculations, we can assume that:
♦ Volume of A = 1 L = 1000 mL
♦ Volume of B = 1 L = 1000 mL
• These assumptions will not affect the results
• Now we can write the steps for finding 'V':
1. 1000 mL of 1 M NaOH solution is kept in a beaker
• This is solution A
• Imagine a 'small cube of side 1 cm' within this solution
2. Volume of that cube = 1 cm3 = 1 cc = 1 mL
• There will be a total of 1000 such small cubes in A
• How many grams of NaOH will be present in that one small cube?
Answer:
(i) 1 M solution means: 1 mole is dissolved in 1000 mL
(ii) A has a volume of 1000 mL
(iii) So 1 mol of NaOH is present in A
(iv) That means 40 g of NaOH is present in A
(v) This 40 g is distributed uniformly among the 1000 small cubes of A
(vi) So one small cube will contain (40 ⁄1000) = 0.04 g
3. Now consider B
(i) Molarity of B is to be: 0.2 M
(ii) 0.2 M solution means: 0.2 mol is dissolved in 1000 mL
(iii) B has a volume of 1000 mL
(iv) So 0.2 mol is to be present in B
(v) That means (0.2 × 40) = 8 g of NaOH is to be present in the 1000 mL of B
4. So we want 8 g of NaOH
(i) Our source is the solution A
(ii) We have seen that, each small cube in A will contain 0.04 g
(iii) Total number of cubes that will contain 8 g = (8⁄0.04) = 200
(iv) So we must take out 200 cubes from A
(v) Each cube is 1 mL. So we can say:
We must take out 200 mL from A
5. We must take this 200 mL in a new beaker
• Add water gradually and bring up the final volume to 1000 mL
• This final 1000 mL solution will be having a molarity of 0.2 M
• If 1500 mL has a molarity of 1 M, then more than 40 g will be present in that 1550 mL
• How much more? Let us find out:
(i) We have: Molarity = (mols⁄1000 mL)
(ii) 1 M means 1 mol is present in 1000 mL
(iii) So in 1 mL, there will be (1⁄1000) = 0.001 mol
(iv) So in 1500 mL, there will be (1500 × 0.001) = 1.5 mols
(v) 1.5 mols = (1.5 × 40) = 60 g
(vi) So there should be 60 g in 1500 mL if it has to be a 1 M solution
7. In that case, the mass of NaOH contained in each small cube = (60 ⁄1500) = 0.04 g
• To obtain 8 g, how many small cubes must be taken out from the 1500 mL?
• The answer is: (8⁄0.04) = 200
• The same value obtained before
8. Thus it is clear that, the volume of the original solution does not matter. It is the molarity that matters
• Similarly, the 'initially assumed volume' of the new solution B also does not matter
• We have to bring up the 200 mL (taken out from A) to 1000 mL by adding water slowly
• So the final volume of B is 1000 mL
Consider the following situation:
(i) A solution is available in a lab. It's volume is 1000 mL
(ii) This solution is named as: Solution A
(iii) The molarity of A is: M1
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution
(vi) This new solution is named as: Solution B
(vii) The molarity of B is to be: M2
(viii) How much is 'V'?
Solution:
1. Let there be m1 grams dissolved in the 1000 mL of A
2. Each 'small cube of 1 cm side' in A will contain $\mathbf\small{\left(\frac{m_1}{1000}\right)}$ grams
3. Now consider B
• Molarity of B is to be: M2
• Let m2 be the mass to be taken out from A
4. Then number of cubes to be taken out = $\mathbf\small{\left(\frac{m_2}{\frac{m_1}{1000}}\right)=\frac{1000m_2}{m_1}}$
• Each cube has a volume of 1 mL
5. So we can write:
Volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,m_2}{m_1}\right)\text{mL}}$
6. Consider the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$
• Both m2 and m1 are in grams
• Dividing both numerator and denominator by 1000 mL, we get:
$\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ = $\mathbf\small{\left[\frac{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}\right]}$
• But $\mathbf\small{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M2
• Similarly, $\mathbf\small{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M1
• So the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ can be written as $\mathbf\small{\left(\frac{M_2}{M_1}\right)}$
7. So the result in (5) will become:
1. Molarity of the original solution of NaOH= 1 M
♦ This is M1
2. Molarity of the new solution = 0.2 M
♦ This is M2
3. So volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,M_2}{M_1}\right)=\left(\frac{1000\times 0.2}{1.0}\right)=200\,\text{mL}}$
4. This 200 mL should be gradually brought up to 1000 mL by adding water
• The final 1000 mL solution thus obtained, will have a molarity of 0.2 M
• On many occasions, reactions are carried out in solutions
• For example, in a reaction, a 'solution of NaOH in water' is one of the reactant
• In such cases, we use any one of the following four methods:
A. Mass percent or weight percent
B. Mole fraction
C. Molarity
D. Molality
We will see each of them in detail:
A. Mass percent
(i) In this method, we first write the masses:
• Mass of the solute (m1)
• Mass of the solvent (m2)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total mass (m1+m2) of the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)^{th}}$ of the total mass is contributed by the solute
• This fraction when expressed as a percentage, gives the mass percent
• So we can write:
Eq.1.3: Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$
An example:
5 g of a substance A is added to 20 grams of a solvent B. What is the mass percent of A?
Solution:
• We have:
Mass percent = $\mathbf\small{\left(\frac{m_1}{m_1+m_2}\right)\times 100}$
♦ Let mass of the substance A be m1
♦ Let mass of the solvent B be m2
• Substituting the values, we get:
Mass percent = $\mathbf\small{\left(\frac{5}{5+20}\right)\times 100=\left(\frac{1}{5}\right)\times 100=20\text{%}}$
B. Mole fraction
(i) In this method, we first write the 'number of moles':
• No. of moles of the solute (n1)
• No. of moles of the solvent (n2)
(ii) We then ask the question:
What is the 'contribution of the solute' towards making up the total number of moles (n1+n2)' in the solution?
(iii) The answer can be written as follows:
$\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)^{th}}$ of the 'total number of moles' is contributed by the solute
(iv) This fraction is called the mole fraction
• Note that, we do not need to convert this fraction into percentage form
• So we can write:
Eq.1.4: Mole fraction = $\mathbf\small{\left(\frac{n_1}{n_1+n_2}\right)}$
C. Molarity
• Detailed notes on molarity can be seen here.
• Based on those notes, we can now derive some interesting results:
• Consider the following situation:
(i) A solution of NaOH is available in a lab
(ii) This solution is named as: Solution A
(iii) The molarity of A is: 1 M
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution of NaOH
(vi) This new solution is named as: Solution B
(vii) The molarity of B should be 0.2 M
(viii) How much is 'V'?
• The above 8 points describe the situation
• But two points seem to be missing:
♦ What is the total volume of A?
♦ What is the total volume of B?
• For calculations, we can assume that:
♦ Volume of A = 1 L = 1000 mL
♦ Volume of B = 1 L = 1000 mL
• These assumptions will not affect the results
• Now we can write the steps for finding 'V':
1. 1000 mL of 1 M NaOH solution is kept in a beaker
• This is solution A
• Imagine a 'small cube of side 1 cm' within this solution
2. Volume of that cube = 1 cm3 = 1 cc = 1 mL
• There will be a total of 1000 such small cubes in A
• How many grams of NaOH will be present in that one small cube?
Answer:
(i) 1 M solution means: 1 mole is dissolved in 1000 mL
(ii) A has a volume of 1000 mL
(iii) So 1 mol of NaOH is present in A
(iv) That means 40 g of NaOH is present in A
(v) This 40 g is distributed uniformly among the 1000 small cubes of A
(vi) So one small cube will contain (40 ⁄1000) = 0.04 g
3. Now consider B
(i) Molarity of B is to be: 0.2 M
(ii) 0.2 M solution means: 0.2 mol is dissolved in 1000 mL
(iii) B has a volume of 1000 mL
(iv) So 0.2 mol is to be present in B
(v) That means (0.2 × 40) = 8 g of NaOH is to be present in the 1000 mL of B
4. So we want 8 g of NaOH
(i) Our source is the solution A
(ii) We have seen that, each small cube in A will contain 0.04 g
(iii) Total number of cubes that will contain 8 g = (8⁄0.04) = 200
(iv) So we must take out 200 cubes from A
(v) Each cube is 1 mL. So we can say:
We must take out 200 mL from A
5. We must take this 200 mL in a new beaker
• Add water gradually and bring up the final volume to 1000 mL
• This final 1000 mL solution will be having a molarity of 0.2 M
6. What if the original 1 M solution A has a volume of 1500 mL?
• How much more? Let us find out:
(i) We have: Molarity = (mols⁄1000 mL)
(ii) 1 M means 1 mol is present in 1000 mL
(iii) So in 1 mL, there will be (1⁄1000) = 0.001 mol
(iv) So in 1500 mL, there will be (1500 × 0.001) = 1.5 mols
(v) 1.5 mols = (1.5 × 40) = 60 g
(vi) So there should be 60 g in 1500 mL if it has to be a 1 M solution
7. In that case, the mass of NaOH contained in each small cube = (60 ⁄1500) = 0.04 g
• To obtain 8 g, how many small cubes must be taken out from the 1500 mL?
• The answer is: (8⁄0.04) = 200
• The same value obtained before
8. Thus it is clear that, the volume of the original solution does not matter. It is the molarity that matters
• Similarly, the 'initially assumed volume' of the new solution B also does not matter
• We have to bring up the 200 mL (taken out from A) to 1000 mL by adding water slowly
• So the final volume of B is 1000 mL
Based on the above discussion, we can derive a general formula:
(i) A solution is available in a lab. It's volume is 1000 mL
(ii) This solution is named as: Solution A
(iii) The molarity of A is: M1
(iv) We want to take out a volume (V mL) from A
(v) Using this 'V mL', we want to make a new solution
(vi) This new solution is named as: Solution B
(vii) The molarity of B is to be: M2
(viii) How much is 'V'?
Solution:
1. Let there be m1 grams dissolved in the 1000 mL of A
2. Each 'small cube of 1 cm side' in A will contain $\mathbf\small{\left(\frac{m_1}{1000}\right)}$ grams
3. Now consider B
• Molarity of B is to be: M2
• Let m2 be the mass to be taken out from A
4. Then number of cubes to be taken out = $\mathbf\small{\left(\frac{m_2}{\frac{m_1}{1000}}\right)=\frac{1000m_2}{m_1}}$
• Each cube has a volume of 1 mL
5. So we can write:
Volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,m_2}{m_1}\right)\text{mL}}$
6. Consider the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$
• Both m2 and m1 are in grams
• Dividing both numerator and denominator by 1000 mL, we get:
$\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ = $\mathbf\small{\left[\frac{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}\right]}$
• But $\mathbf\small{\left(\frac{m_2}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M2
• Similarly, $\mathbf\small{\left(\frac{m_1}{1000}\right)\left(\frac{\text{grams}}{\text{Litre}}\right)}$ = The molarity M1
• So the ratio $\mathbf\small{\left(\frac{m_2}{m_1}\right)}$ can be written as $\mathbf\small{\left(\frac{M_2}{M_1}\right)}$
7. So the result in (5) will become:
Eq.1.5:
Volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,M_2}{M_1}\right)\text{mL}}$
Let us apply this equation to our present case:
♦ This is M1
2. Molarity of the new solution = 0.2 M
♦ This is M2
3. So volume to be taken out from the original solution = $\mathbf\small{\left(\frac{1000\,M_2}{M_1}\right)=\left(\frac{1000\times 0.2}{1.0}\right)=200\,\text{mL}}$
4. This 200 mL should be gradually brought up to 1000 mL by adding water
• The final 1000 mL solution thus obtained, will have a molarity of 0.2 M
Solved example 1.20
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution
Solution:
1. Assume that, the 250 mL is taken out from a solution A of volume 1000 mL
2. Then we can write:
250 mL of A has 4 g
3. So 1 mL of A has (4⁄250) g
4. So 1000 mL of A has [(4⁄250) × 1000] = 16 g
5. So 16 g is dissolved in 1000 mL
6. We can write:
4 g dissolved in 250 mL
is equivalent to
16 g dissolved in 1000 mL
7. One mole of NaOH = 40 g
• So 1 g = (1⁄40) mole
• So 16 g = [16 × (1⁄40)] = 0.4 mole
8. We can write:
• 0.4 mole is dissolved in 1000 mL
• But 'no. of moles dissolved in 1000 mL' is the molarity
9. So we get:
4 g dissolved in 250 mL
is equivalent to
a molarity of 0.4 M
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution
Solution:
1. Assume that, the 250 mL is taken out from a solution A of volume 1000 mL
2. Then we can write:
250 mL of A has 4 g
3. So 1 mL of A has (4⁄250) g
4. So 1000 mL of A has [(4⁄250) × 1000] = 16 g
5. So 16 g is dissolved in 1000 mL
6. We can write:
4 g dissolved in 250 mL
is equivalent to
16 g dissolved in 1000 mL
7. One mole of NaOH = 40 g
• So 1 g = (1⁄40) mole
• So 16 g = [16 × (1⁄40)] = 0.4 mole
8. We can write:
• 0.4 mole is dissolved in 1000 mL
• But 'no. of moles dissolved in 1000 mL' is the molarity
9. So we get:
4 g dissolved in 250 mL
is equivalent to
a molarity of 0.4 M
Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent
• When the temperature increases, the volume also increases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator increases
♦ So molarity decreases
• When the temperature decreases, the volume also decreases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator decreases
♦ So molarity increases
• When the temperature increases, the volume also increases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator increases
♦ So molarity decreases
• When the temperature decreases, the volume also decreases
♦ The number of moles which is in the numerator, remains the same
♦ The volume which is in the denominator decreases
♦ So molarity increases
D. Molality
In this method we first write two items:
(i) The number of moles of the solute
(ii) The mass of the solvent in kg
• Their ratio is called molality. It's symbol is m
• So we get:
Eq.1.6: Molality (m) = $\mathbf\small{\frac{\text{No. of moles of solute}}{\text{mass of solvent in kg}}}$
• So in effect, m is the 'number of moles of the solute' present in 'each 1 kg' of the solvent
Solved example 1.21
The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate the molality of the solution
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.25 g mL-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.25 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.25 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1250\, \text{g}}$
3. This 1250 g is the total mass of the solution
• So we can write:
1250 g = Mass of the NaCl + Mass of the solvent
4. Molarity of the solution is given as 3 M
• So every 1000 mL of the solution will contain 3 moles of NaCl
5. Three moles of NaCl = (3 × 58.5) = 175.5 g
• So from the result in (3), we get:
1250 = 175.5 + Mass of the solvent
⇒ Mass of the solvent = (1250-175.5) = 1074.5 g = 1.0745 kg
6. So, for every 1.0745 kg, there are 3 moles of NaCl
• Thus we get:
Molatity = No. of moles per kg = $\mathbf\small{\frac{\text{3 (mole)}}{\text{1.0745 (kg)}}=2.79\, \text{mol kg}^{-1}}$
In this method we first write two items:
(i) The number of moles of the solute
(ii) The mass of the solvent in kg
• Their ratio is called molality. It's symbol is m
• So we get:
Eq.1.6: Molality (m) = $\mathbf\small{\frac{\text{No. of moles of solute}}{\text{mass of solvent in kg}}}$
• So in effect, m is the 'number of moles of the solute' present in 'each 1 kg' of the solvent
Solved example 1.21
The density of 3 M solution of NaCl is 1.25 g mL-1. Calculate the molality of the solution
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.25 g mL-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.25 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.25 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1250\, \text{g}}$
3. This 1250 g is the total mass of the solution
• So we can write:
1250 g = Mass of the NaCl + Mass of the solvent
4. Molarity of the solution is given as 3 M
• So every 1000 mL of the solution will contain 3 moles of NaCl
5. Three moles of NaCl = (3 × 58.5) = 175.5 g
• So from the result in (3), we get:
1250 = 175.5 + Mass of the solvent
⇒ Mass of the solvent = (1250-175.5) = 1074.5 g = 1.0745 kg
6. So, for every 1.0745 kg, there are 3 moles of NaCl
• Thus we get:
Molatity = No. of moles per kg = $\mathbf\small{\frac{\text{3 (mole)}}{\text{1.0745 (kg)}}=2.79\, \text{mol kg}^{-1}}$
We have completed a discussion on the basics of the four methods. Now we will see some more solved examples
Solved example 1.22
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
Solution:
1. We want a 0.375 M solution of sodium acetate
• So 1000 mL of that solution should contain 0.375 moles
2. One mole of sodium acetate is 82.0245 g
• So 0.375 mol = (0.375× 82.0245) = 30.7592 g
3. This 30.7592 g of sodium acetate is to be dissolved in 1000 mL
• So for 500 mL, (30.7592⁄2) = 15.3796 g of sodium acetate will be sufficient
Solved example 1.23
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL-1 and the mass per cent of nitric acid in it being 69%.
Solution:
1. Consider 1000 mL volume of the sample
• What ever be the volume, the density will not change
• So the 1000 mL of the sample will have a density of 1.41 g mL-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.41 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.41 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1410\, \text{g}}$
3. This 1410 g is the total mass of the solution
• So we can write:
1410 g = Mass of the nitric acid + Mass of the solvent
4. Given that, mass percent of nitric acid is 69%
• That means, 69% of 1410 g is the 'mass of nitric acid'
• So we get:
Mass of nitric acid in 1000 mL solution = (1410 × 0.69) = 972.9 g
5. One mol of nitric acid (HNO3) = (1 × 1.008) + (1 × 14) + (3 × 16) = 63.01 g
• So 972.9 g = (972.9⁄63.01) = 15.44 mols
6. So we get:
• 15.44 mols will be present in 1 liter
• So molarity of the solution is 15.44 M
Solved example 1.24
What is the molarity of H2SO4 solution that has a density of 1.84 g/cc at 35o c and contains 98% by weight?
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.84 g mL-1
(∵ 1 cc = 1 mL)
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.84 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.84 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1840\, \text{g}}$
3. This 1840 g is the total mass of the solution
• Given that 98% of the total mass is H2SO4
• So we can write:
• Mass of H2SO4 = (1840 × 0.98)
4. Molar mass of H2SO4 = (2 ×1) + (1 × 32) + (4 × 16) = 98 g
• So number of moles in the 1000 mL = $\mathbf\small{\frac{1840 \times 0.98}{98}}$ = 18.40 mols
5. But 'number of mols in 1000 mL' is molarity
• So we can write:
Molarity of the given solution is 18.4 M
Solved example 1.22
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
Solution:
1. We want a 0.375 M solution of sodium acetate
• So 1000 mL of that solution should contain 0.375 moles
2. One mole of sodium acetate is 82.0245 g
• So 0.375 mol = (0.375× 82.0245) = 30.7592 g
3. This 30.7592 g of sodium acetate is to be dissolved in 1000 mL
• So for 500 mL, (30.7592⁄2) = 15.3796 g of sodium acetate will be sufficient
Solved example 1.23
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL-1 and the mass per cent of nitric acid in it being 69%.
Solution:
1. Consider 1000 mL volume of the sample
• What ever be the volume, the density will not change
• So the 1000 mL of the sample will have a density of 1.41 g mL-1.
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.41 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.41 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1410\, \text{g}}$
3. This 1410 g is the total mass of the solution
• So we can write:
1410 g = Mass of the nitric acid + Mass of the solvent
4. Given that, mass percent of nitric acid is 69%
• That means, 69% of 1410 g is the 'mass of nitric acid'
• So we get:
Mass of nitric acid in 1000 mL solution = (1410 × 0.69) = 972.9 g
5. One mol of nitric acid (HNO3) = (1 × 1.008) + (1 × 14) + (3 × 16) = 63.01 g
• So 972.9 g = (972.9⁄63.01) = 15.44 mols
6. So we get:
• 15.44 mols will be present in 1 liter
• So molarity of the solution is 15.44 M
Solved example 1.24
What is the molarity of H2SO4 solution that has a density of 1.84 g/cc at 35o c and contains 98% by weight?
Solution:
1. Consider 1000 mL volume of the given solution
• What ever be the volume, the density will not change
• So the 1000 mL of the given solution will have a density of 1.84 g mL-1
(∵ 1 cc = 1 mL)
2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$
$\mathbf\small{\Rightarrow\text{1.84 (g mL)}^{-1}=\frac{\text{Total mass of the solution}}{\text{1000 (mL)}}}$
$\mathbf\small{\Rightarrow \text{Total mass of the solution}= \text{1.84 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1840\, \text{g}}$
3. This 1840 g is the total mass of the solution
• Given that 98% of the total mass is H2SO4
• So we can write:
• Mass of H2SO4 = (1840 × 0.98)
4. Molar mass of H2SO4 = (2 ×1) + (1 × 32) + (4 × 16) = 98 g
• So number of moles in the 1000 mL = $\mathbf\small{\frac{1840 \times 0.98}{98}}$ = 18.40 mols
5. But 'number of mols in 1000 mL' is molarity
• So we can write:
Molarity of the given solution is 18.4 M
In the next section, we will see a few more solved examples
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