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:
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:
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
• 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
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
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