Chapter 1.6 - Molarity - Solved Examples

In the previous section, we saw the basics about mass percent, mole fraction, molarity and molality. We also saw some solved examples. In this section we will see a few more solved examples

Solved example 1.25
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if its 20 g are dissolved in enough water to make a final volume up to 2L?
Solution:
1. Molar mass of sugar = (12 ×12.01) + (1.008 × 22) + (11 × 16) = 342.3 g
⇒ 1 mol of sugar =342.3 g
⇒ 1 g of sugar = (¹⁄_342.3) mol 
⇒ 20 g of sugar = [20 × (¹⁄_342.3)] = (²⁰⁄_342.3) mol
2. (²⁰⁄_342.3) mol is present in 2L
⇒ Number of moles present in 1 L = Molarity = [¹⁄₂ × (²⁰⁄_342.3)] = 0.0292 M

Solved example 1.26
If the density of methanol is 0.793 kg L^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Solution:
1. Density is given as 0.793 kg L^-1
So it is clear that, in 1 L of methanol, there will be 0.793 kg of methanol
2. The molar mass of methanol (CH₃OH) is 32.04 g = 0.03204 kg
⇒ Number of moles present in 0.793 kg = (^0.793⁄_0.03204) = 24.75 mols
⇒ 24.75 mols of methanol is present in it's 1 L volume
3. We want a 0.25 M solution of methanol
So 0.25 mols should be present in 1 L
4. So in 2.5 L, (2.5 × 0.25) = 0.625 mols must be present
5. From the result in (2), we have:
24.75 mols of methanol is present in it's 1 L volume
⇒ 1 mol of methane will be present in it's (¹⁄_24.75) L volume
6. So 0.625 mols will be present in it's [0.625 × (¹⁄_24.75)] = 0.02525 L volume
• We have: 0.02525 L = 25.2 mL
• So we can write:
25.2 mL of methanol is required to make 2.5 L of it's 0.25 M solution

Solved example 1.27
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Solution:
1. Molar mass of = 106 g
⇒ 0.5 mol Na₂CO₃ = 53 g
2. 0.5 M Na₂CO₃ means that 53 g of Na₂CO₃ is dissolved in 1 L solution of Na₂CO₃.

Solved example 1.28
If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution:
1. The balanced equation is: 2H₂ (g) + O₂ (g) → 2H₂O (g)
2. Hydrogen and oxygen are in the ratio 2:1
• So 5 volumes of oxygen i sufficient for 10 volumes of hydrogen
3. Hydrogen and water vapour are in the ratio 1 : 1
• So in this reaction, 10 volumes of water vapour will be produced

Solved example 1.29
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Solution:
1. Given that, 3.38 g of CO₂ is produced
2. Consider any sample of CO₂ 
• Using Eq.1.1, we can find the % mass of carbon in it
We have:
Percentage mass of an element in a pure sample of any of it's compound 
= $\mathbf\small{\left(\frac{(nA)_{\rm{Element}} \times (GAM)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)\times 100}$
3. Substituting the values, we get:
Percentage mass of carbon in a pure sample of CO₂
= $\mathbf\small{\left(\frac{1 \times 12}{44}\right)\times 100=\frac{300}{11}\text{%}}$
4. So (³⁰⁰⁄₁₁) % of 3.38 g is C
⇒ Mass of carbon in the 3.38 g of CO₂ = (3.38 × ³⁄₁₁) = 0.921818 g
5. The only source of this C is the gas fuel
So we can write:
Mass of C in the gas fuel = 0.921818 g
6. Consider any sample of H₂O
• Using Eq.1.1 above, we can find the % mass of H in it
7. Substituting the values, we get:
Percentage mass of H in a pure sample of H₂O
= $\mathbf\small{\left(\frac{2 \times 1}{18}\right)\times 100=\frac{100}{9}\text{%}}$
8. So (¹⁰⁰⁄₉) % of 0.18 g is H
⇒ Mass of H in the 0.690 g of H₂O = (0.690 × ¹⁄₉) = 0.07667 g
9. The only source of this H is the gas fuel
So we can write:
Mass of H in the gas fuel = 0.07667 g
10. Total mass of C and H in the gas fuel = (0.921818 + 0.07667) = 0.998488 g
11. Now we can find the percentages:
% mass of C in the gas fuel = (^0.921818⁄_0.998488) × 100 = 92.32% 
% mass of H in the gas fuel = (^0.07667⁄_0.998488) × 100 = 7.68%
12. Now we use Eq.1.2:

$\mathbf\small{\left(\frac{(nA)_{\rm{Element}}}{(GMM)_{\rm{Compound}}}\right)=\frac{\text{% of the element}}{\text{GAM of the element} \times 100}}$
13. We have to find the GMM of the compound
• 10 L weighs 11.6 g
• So 1 L weighs 1.16 g
• So 22.7 L weighs (22.7 × 1.16) = 26.332 g
• Thus we get: GMM of the compound = 26.332 g
• This is the answer for part (ii)
14. Substituting the values for C, we get:

$\mathbf\small{\left(\frac{(nA)_{\rm{C}}}{26.332}\right)=\frac{92.32}{12 \times 100}}$

⇒ (nA)_C = 2.026 ≈ 2

• Substituting the values for H, we get:

$\mathbf\small{\left(\frac{(nA)_{\rm{H}}}{26.332}\right)=\frac{\frac{7.68}{3}}{1 \times 100}}$

⇒ (nA)_H = 2.022 ≈ 2

15. So the molecular formula is: C₂H₂
• This is the answer for part (iii)
16. Carbon and hydrogen atoms are in the ratio 2 : 2
Reducing to the simplest form, this ratio is 1 : 1
So the empirical formula is C₁H₁.
• This is the answer for part (i)

Solved example 1.30
Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction:
CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂(g) + H₂O (l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Solution:
1. 0.75 M HCl means that 0.75 mol is dissolved in 1 L of HCL solution
• 1 mol HCl = 36.5 g
• So 0.75 mol = (0.75 × 36.5) = 27.375 g
⇒ 27.375 g of HCl is dissolved in 1000 mL of HCl solution
⇒ In 1 mL of that solution, (^27.375⁄₁₀₀₀) g of HCl is dissolved
⇒ In 25 mL of that solution, [25×(^27.375⁄₁₀₀₀)] = 0.6844 g of HCl is dissolved   
2. 1 mol HCl = 36.5 g
⇒ 1 g HCl = (¹⁄_36.5) mol
⇒ 0.6844 g HCl = [0.6844 × (¹⁄_36.5)] = 0.0187 mol
3. CaCO₃ and HCl are in the ratio 1 : 2
• So number of moles of CaCO₃ required = (^0.0187⁄₂) = 0.00935 mol
4. one mol CaCO₃ is 100.08 g
• So 0.00935 mol = (0.00935 × 100.08) = 0.935 g

Solved example 1.31
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4 HCl (aq) + MnO₂(s) ⟶ 2H₂O (l) + MnCl₂(aq) + Cl₂ (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution:
1. Molar mass of MnO₂ is 86.94 g
⇒ 1 mol MnO₂ = 86.94 g
⇒ 1 g MnO₂ = (¹⁄_86.94) mol

⇒ 5.0 g MnO₂ = [5×(¹⁄_86.94)] = 0.0575 mol

2. In the balanced equation, HCl and MnO₂ are in the ratio 4 : 1

• So (4 × 0.0575) = 0.23 mol of HCl will take part in the reaction

3. Molar mass of HCl is 36.5 g

⇒ 1 mol of HCl = 36.5 g

⇒ 0.23 mol of HCl = (0.23 ×36.5) = 8.395 g

Solved example 1.32

What is the molarity of liquid HCl that has a density of 1.17 g/cc?

Solution:

1. Consider 1000 mL volume of the given liquid

• What ever be the volume, the density will not change

• So the 1000 mL of the given liquid will have a density of 1.17 g mL^-1

(∵ 1 cc = 1 mL)

2. We have: $\mathbf\small{\text{Density}=\frac{\text{Mass}}{\text{Volume}}}$

$\mathbf\small{\Rightarrow\text{1.17 (g mL)}^{-1}=\frac{\text{Total mass of the liquid}}{\text{1000 (mL)}}}$

$\mathbf\small{\Rightarrow \text{Total mass of the liquid}= \text{1.17 (g mL)}^{-1}\times 1000\,\text{(mL)}= 1170\, \text{g}}$

3. This 1170 g is the total mass of the liquid

4. Molar mass of HCl = 36.5 g

• So number of moles in the 1000 mL = $\mathbf\small{\frac{1170}{36.5}}$ = 32.05 mols

5. But 'number of mols in 1000 mL' is molarity

• So we can write:

Molarity of the given liquid HCl = 32.05 M

Solved example 1.33

What will be the molality of a solution made by dissolving 4 g of a salt having molar mass 40 g in 250 g of water?

Solution:

1. No. of mols of the salt = (⁴⁄₄₀) = 0.1 mol

2. So 0.1 mol of the solute is present in 250 g of the solvent

⇒ 0.1 mol of the solute is present in 0.25 kg of the solvent

3. So, for 1 kg og the solvent, (^0.1⁄_0.25) = 0.4 mols

4. But number of mols of the solute per kg of the solvent is molality

So we can write:

Molality of the solution will be 0.4

Solved example 1.34

12 g of Mg reacts with hydrochloric acid to give hydrogen gas. What will be the volume of hydrogen gas produced at STP? 

Solution:

1. The balanced equation is: Mg (s) + 2HCl (aq) ⟶ MgCl₂(aq) + H₂ (g)
2. Molar mass of Mg is 24 g
⇒ 1 mol of Mg = 24 g
⇒ 1 g of Mg = (¹⁄₂₄) mol
⇒ 12 g of Mg = [12 × (¹⁄₂₄)] = 0.5 mol
3. In the balanced equation, Mg and H₂ are in the ratio 1 : 1
• So 0.5 mol of H₂ will be produced 
4. Volume of 0.5 mol H₂ at STP is (^22.7 ⁄₂) = 11.35 L

Solved example 1.35
Calculate the amount of zinc required to produce 227 mL of H₂ at STP on treatment with dilute H₂SO₄ 
Solution:
1. The balanced equation is: Zn (s) + H₂SO₄ (aq) ⟶ ZnSO₄ (aq) + H₂ (g)
2. The molar mass of Zn is 65 g
So 1 mol of Zn is 65 g
3. In the balanced equation, Zn and H₂ are in the ratio 1 : 1
• So 1 mol of Zn will give 1 mol of H₂.
4. But 1 mol of H₂ is 22.7 L
• So we get:
1 mol of Zn will give 22700 mL of H₂.
⇒ For 1 mL of H₂, (¹⁄₂₂₇₀₀) mol of Zn is required
⇒ For 227 mL of H₂, [227 × (¹⁄₂₂₇₀₀)] = 0.01 mol of Zn is required
5. 0.01 mol of Zn = 0.65 g

---

In the next chapter, we will see structure of atom

PREVIOUS           CONTENTS          NEXT


Copyright©2019 Higher Secondary Chemistry. blogspot.in - All Rights Reserved

Chapter 1.5 - Molarity of Solutions

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

---

6. What if the original 1 M solution A has a volume of 1500 mL?  

• 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

---

Based on the above discussion, we can derive a general formula:

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:

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:

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

---

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

---

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

---

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 = $\mathbf\small{\frac{\text{3 (mole)}}{\text{1.0745 (kg)}}=2.79\, \text{mol kg}^{-1}}$

---

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 35^o 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

PREVIOUS           CONTENTS          NEXT


Copyright©2019 Higher Secondary Chemistry. blogspot.in - All Rights Reserved