Solved Examples

Solved Examples

This section contains 30+ worked problems covering all the major topics of this chapter — from basic unit conversions and significant figures to advanced stoichiometry and limiting reagent problems. These are the types of questions you'll encounter in Board exams, JEE Main, and NEET.

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Example 1: Unit Conversion — SI Prefixes

Convert 25.6 mg to kg.

Solution:

1. $$25.6 \text{ mg} = 25.6 \times 10^{-3} \text{ g}$$
2. $$= 25.6 \times 10^{-3} \times 10^{-3} \text{ kg}$$
3. $$= 25.6 \times 10^{-6} \text{ kg}$$
4. $$= 2.56 \times 10^{-5} \text{ kg}$$

Final Answer: $$2.56 \times 10^{-5} \text{ kg}$$

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Example 2: Temperature Conversion

Convert 373 K to °C and °F.

Solution:

1. $$^\circ \text{C} = K - 273.15 = 373 - 273.15 = 99.85^\circ \text{C}$$
2. $$^\circ \text{F} = \frac{9}{5} \times 99.85 + 32 = 179.73 + 32 = 211.73^\circ \text{F}$$

Final Answer: $$99.85^\circ \text{C} \text{ and } 211.73^\circ \text{F}$$

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Example 3: Significant Figures in Addition

Add: 12.11 + 18.0 + 1.012

Solution:

1. Raw sum = 31.122
2. In addition, the result should have the same number of decimal places as the measurement with the fewest decimal places.
3. 18.0 has 1 decimal place.
4. Therefore, $31.122 \to 31.1$

Final Answer: $31.1$

Example 4: Significant Figures in Multiplication

Calculate $2.5 \times 1.25$.

Solution:

• Raw product = 3.125
• In multiplication, the result should have the same number of significant figures as the measurement with fewest significant figures
• 2.5 has 2 significant figures (fewest)
• Answer: 3.1 (rounded to 2 significant figures)

Final Answer: 3.1

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Example 5: Scientific Notation

Express the following in scientific notation: (a) 0.0048 (b) 234000 (c) 8008

Solution:

• (a) $0.0048 = 4.8 \times 10^{-3}$
• (b) $234000 = 2.34 \times 10^{5}$
• (c) $8008 = 8.008 \times 10^{3}$

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Example 6: Dimensional Analysis

Convert a speed of 90 km/h to m/s.

Solution: $$90 \text{ km/h} = 90 \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{90000}{3600} = 25 \text{ m/s}$$

Shortcut: Divide by 3.6 → $90/3.6 = 25$ m/s

Final Answer: 25 m/s

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Example 7: Density Calculation

A piece of metal weighs 25 g and has a volume of 5 cm³. Calculate its density in SI units (kg/m³).

Solution:

1. Density = $\frac{25 \text{ g}}{5 \text{ cm}^3}$ = 5 g/cm³
2. Convert: $5 \text{ g/cm}^3 = 5 \times \frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 5 \times 10^{3} \text{ kg/m}^3 = 5000 \text{ kg/m}^3$

Final Answer: 5000 kg/m³

Example 8: Law of Conservation of Mass

When 4.2 g of $\text{NaHCO}_3$ is heated, it decomposes into $\text{Na}_2\text{CO}_3$, $\text{H}_2\text{O}$, and $\text{CO}_2$. Calculate the total mass of products.

$$2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$$

Solution: By the Law of Conservation of Mass, the total mass of products = total mass of reactants = 4.2 g.

We can verify: Moles of $\text{NaHCO}_3$ = 4.2/84 = 0.05 mol.

• $\text{Na}_2\text{CO}_3$: 0.025 mol = 0.025 × 106 = 2.65 g
• $\text{H}_2\text{O}$: 0.025 mol = 0.025 × 18 = 0.45 g
• $\text{CO}_2$: 0.025 mol = 0.025 × 44 = 1.1 g
• Total = 2.65 + 0.45 + 1.1 = 4.2 g ✓

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Example 9: Law of Multiple Proportions

Carbon and oxygen form CO and $\text{CO}_2$. In CO, 3 g of carbon combines with 4 g of oxygen. In $\text{CO}_2$, 3 g of carbon combines with 8 g of oxygen. Verify the law of multiple proportions.

Solution: Fixing mass of carbon at 3 g:

• Oxygen in CO = 4 g
• Oxygen in $\text{CO}_2$ = 8 g
• Ratio = 4:8 = 1:2 — a simple whole number ratio ✓

This confirms the Law of Multiple Proportions.

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Example 10: Average Atomic Mass

Boron exists as two isotopes: $^{10}\text{B}$ (19.9%) and $^{11}\text{B}$ (80.1%). Calculate the average atomic mass of boron.

Solution: $$\text{Average} = \frac{19.9 \times 10 + 80.1 \times 11}{100} = \frac{199 + 881.1}{100} = \frac{1080.1}{100} = 10.801 \text{ u}$$

Final Answer: 10.801 u (which matches the periodic table value!)

Example 11: Molecular Mass Calculation

Calculate the molecular mass of calcium hydroxide, $\text{Ca(OH)}_2$. (Atomic masses: Ca = 40, O = 16, H = 1)

Solution: $$M = 40 + 2(16 + 1) = 40 + 34 = 74 \text{ u}$$

Final Answer: 74 u (or 74 g/mol as molar mass)

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Example 12: Formula Mass of Ionic Compound

Calculate the formula mass of sodium carbonate ($\text{Na}_2\text{CO}_3$). (Na = 23, C = 12, O = 16)

Solution: $$\text{Formula mass} = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 \text{ u}$$

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Example 13: Mass of a Single Atom

Calculate the mass of a single atom of gold (Au). (Atomic mass = 197 u)

Solution: $$m = \frac{197}{6.022 \times 10^{23}} = 3.271 \times 10^{-22} \text{ g}$$

Final Answer: $3.271 \times 10^{-22}$ g

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Example 14: Moles from Mass

Calculate the number of moles of potassium hydroxide (KOH) in 280 g. (K = 39, O = 16, H = 1)

Solution: Molar mass of KOH = 39 + 16 + 1 = 56 g/mol $$n = \frac{280}{56} = 5 \text{ mol}$$

Final Answer: 5 moles

Example 15: Moles to Particles

How many formula units are present in 11.7 g of NaCl? (M = 58.5 g/mol)

Solution:

1. Moles = 11.7/58.5 = 0.2 mol
2. Formula units = $0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$

Bonus: How many Na⁺ ions? Also $1.2044 \times 10^{23}$ (one per formula unit). How many Cl⁻ ions? Also $1.2044 \times 10^{23}$. Total ions = $2.4088 \times 10^{23}$.

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Example 16: Volume of Gas at STP

What volume of oxygen gas at STP is needed to burn 1 mole of ethane ($\text{C}_2\text{H}_6$)?

$$2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$$

Solution:

1. 2 mol $\text{C}_2\text{H}_6$ requires 7 mol $\text{O}_2$
2. 1 mol $\text{C}_2\text{H}_6$ requires 3.5 mol $\text{O}_2$
3. Volume = $3.5 \times 22.4 = 78.4$ L

Final Answer: 78.4 L at STP

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Example 17: Number of Atoms in a Compound

How many atoms of each type are present in 9.8 g of $\text{H}_2\text{SO}_4$?

Solution:

1. Moles of $\text{H}_2\text{SO}_4$ = 9.8/98 = 0.1 mol
2. Molecules = $0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$
3. H atoms: $2 \times 6.022 \times 10^{22} = 1.2044 \times 10^{23}$
4. S atoms: $1 \times 6.022 \times 10^{22} = 6.022 \times 10^{22}$
5. O atoms: $4 \times 6.022 \times 10^{22} = 2.4088 \times 10^{23}$
6. Total atoms: $1.2044 \times 10^{23} + 6.022 \times 10^{22} + 2.4088 \times 10^{23} = 4.2154 \times 10^{23}$

Example 18: Percentage Composition

Calculate the percentage of nitrogen in ammonium nitrate ($\text{NH}_4\text{NO}_3$).

Solution:

1. Molar mass = $14 + 4 + 14 + 48 = 80 \text{ g/mol}$
2. Mass of nitrogen = $28 \text{ g}$
3. $$\%N = \frac{28}{80} \times 100 = 35\%$$

Final Answer: $35\%$

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Example 19: Empirical Formula from Mass Data

A compound is found to contain 2.04 g of sodium, 2.65 g of carbon, and 7.06 g of oxygen. What is its empirical formula? (Na = 23, C = 12, O = 16)

Solution:

1. Convert masses to moles:

Element	Mass (g)	Atomic Mass	Moles
Na	2.04	23	$$\frac{2.04}{23} = 0.0887$$
C	2.65	12	$$\frac{2.65}{12} = 0.2208$$
O	7.06	16	$$\frac{7.06}{16} = 0.4413$$

2. Divide by the smallest value (0.0887):

• Na: $\frac{0.0887}{0.0887} = 1$
• C: $\frac{0.2208}{0.0887} \approx 2.49 \approx 2.5$
• O: $\frac{0.4413}{0.0887} \approx 4.98 \approx 5$

2. Ratio is approximately: $$1 : 2.5 : 5$$

3. Multiply by 2 to convert into whole numbers: $$2 : 5 : 10$$

4. Therefore, the empirical formula is: $$\text{Na}_2\text{C}_5\text{O}_{10}$$

Final Answer: $$\text{Na}_2\text{C}_5\text{O}_{10}$$

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Example 20: Molecular Formula from Molar Mass

A hydrocarbon contains 85.7% carbon and 14.3% hydrogen. Its molar mass is 42 g/mol. Find the molecular formula.

Solution:

1. Assume 100 g of compound.

Element	Mass (g)	Atomic Mass	Moles	Ratio
C	85.7	12	7.142	1
H	14.3	1	14.3	2

2. Empirical formula = $\text{CH}_2$
3. Empirical formula mass = $14$
4. $n = \frac{42}{14} = 3$
5. Molecular formula = $(\text{CH}_2)_3 = \text{C}_3\text{H}_6$

Final Answer: $\text{C}_3\text{H}_6$

Example 21: Stoichiometry — Mass to Mass

How many grams of hydrogen are produced when 13 g of zinc reacts with excess hydrochloric acid?

$$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$$

Solution:

1. Moles of Zn = 13/65 = 0.2 mol
2. 1 mol Zn → 1 mol $\text{H}_2$
3. $\text{H}_2$ = 0.2 mol = 0.2 × 2 = 0.4 g

Final Answer: 0.4 g

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Example 22: Stoichiometry — Volume to Volume (Gases)

What volume of oxygen is needed to burn 100 mL of methane (both at STP)?

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

Solution: By Gay Lussac's law, volume ratio = mole ratio (at same T, P). 1 vol $\text{CH}_4$ : 2 vol $\text{O}_2$ So 100 mL $\text{CH}_4$ needs 200 mL $\text{O}_2$.

Final Answer: 200 mL

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Example 23: Limiting Reagent with Excess Calculation

2.8 g of nitrogen reacts with 1 g of hydrogen to form ammonia. Find the mass of ammonia formed and the mass of excess reagent left over.

$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

Solution:

1. Moles: $\text{N}_2$ = 2.8/28 = 0.1 mol; $\text{H}_2$ = 1/2 = 0.5 mol
2. Required $\text{H}_2$ for 0.1 mol $\text{N}_2$: $0.1 \times 3 = 0.3$ mol
3. Available $\text{H}_2$ = 0.5 mol > 0.3 mol → $\text{N}_2$ is limiting
4. $\text{NH}_3$ formed: $0.1 \times 2 = 0.2$ mol = $0.2 \times 17 = 3.4$ g
5. $\text{H}_2$ consumed: 0.3 mol = 0.6 g
6. $\text{H}_2$ remaining: 1 − 0.6 = 0.4 g

Final Answer: 3.4 g of $\text{NH}_3$; 0.4 g of $\text{H}_2$ in excess.

Example 24: Molarity from Mass and Volume

What is the molarity of a solution containing 4.0 g of NaOH in 250 mL of solution? (M of NaOH = 40)

Solution:

1. Moles = 4.0/40 = 0.1 mol
2. Volume = 250 mL = 0.25 L
3. Molarity = 0.1/0.25 = 0.4 M

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Example 25: Preparing a Solution of Given Molarity

How would you prepare 500 mL of 0.2 M sodium carbonate ($\text{Na}_2\text{CO}_3$) solution? (M = 106 g/mol)

Solution:

1. Moles needed = $M \times V$ = $0.2 \times 0.5$ = 0.1 mol
2. Mass = $0.1 \times 106$ = 10.6 g

Procedure: Dissolve 10.6 g of $\text{Na}_2\text{CO}_3$ in some water and make up the total volume to 500 mL.

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Example 26: Dilution Problem

What volume of 12 M HCl is needed to prepare 1 L of 0.5 M HCl?

Solution: $$M_1V_1 = M_2V_2$$ $$12 \times V_1 = 0.5 \times 1000$$ $$V_1 = \frac{500}{12} = 41.67 \text{ mL}$$

Procedure: Take 41.67 mL of concentrated (12 M) HCl and dilute to 1000 mL.

Safety Note: Always add acid to water, never the reverse! ("Do as you oughta — add acid to water.")

Example 27: Mole Fraction Calculation

Calculate the mole fraction of ethanol in a solution containing 46 g of ethanol ($\text{C}_2\text{H}_5\text{OH}$, M = 46) and 72 g of water.

Solution:

1. Moles of ethanol = 46/46 = 1 mol
2. Moles of water = 72/18 = 4 mol
3. Total moles = 5 mol

$$x_{\text{ethanol}} = \frac{1}{5} = 0.2$$ $$x_{\text{water}} = \frac{4}{5} = 0.8$$

Check: 0.2 + 0.8 = 1 ✓

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Example 28: Molality Calculation

Calculate the molality of a solution containing 10 g of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, M = 180) in 200 g of water.

Solution:

1. Moles of glucose = 10/180 = 0.0556 mol
2. Mass of solvent = 200 g = 0.2 kg
3. Molality = 0.0556/0.2 = 0.278 m

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Example 29: Converting Mass % to Molarity

A 20% (w/w) solution of $\text{H}_2\text{SO}_4$ has a density of 1.14 g/mL. Find its molarity.

Solution:

1. Consider 1 L of solution: mass = $1000 \times 1.14$ = 1140 g
2. Mass of $\text{H}_2\text{SO}_4$ = 20% of 1140 = 228 g
3. Moles = 228/98 = 2.327 mol
4. Molarity = 2.327/1 = 2.327 M

Example 30: Multi-step Stoichiometry

Consider: $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$

If 24.5 g of $\text{KClO}_3$ (M = 122.5) decomposes: (a) What mass of $\text{O}_2$ is produced? (b) What volume does this $\text{O}_2$ occupy at STP? (c) How many molecules of $\text{O}_2$ are produced?

Solution:

1. Moles of $\text{KClO}_3$ = 24.5/122.5 = 0.2 mol

(a) Moles of $\text{O}_2$ = $\frac{3}{2} \times 0.2$ = 0.3 mol Mass = $0.3 \times 32$ = 9.6 g

(b) Volume at STP = $0.3 \times 22.4$ = 6.72 L

(c) Molecules = $0.3 \times 6.022 \times 10^{23}$ = $1.807 \times 10^{23}$

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Example 31: Stoichiometry with Concentration

How many mL of 0.5 M $\text{H}_2\text{SO}_4$ are required to dissolve 0.5 g of copper(II) carbonate?

$$\text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2$$

Solution:

1. Molar mass of $\text{CuCO}_3$ = 63.5 + 12 + 48 = 123.5 g/mol
2. Moles of $\text{CuCO}_3$ = 0.5/123.5 = 0.00405 mol
3. Mole ratio 1:1 → Moles of $\text{H}_2\text{SO}_4$ = 0.00405 mol
4. Volume = $\frac{0.00405}{0.5}$ = 0.0081 L = 8.1 mL

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Example 32: Comparing Masses of Equal Moles

Arrange the following in order of increasing mass: 1 mol Na, 1 mol Fe, 1 mol Al, 1 mol Ag.

Solution:

• 1 mol Na = 23 g
• 1 mol Al = 27 g
• 1 mol Fe = 56 g
• 1 mol Ag = 108 g

Order: Na (23 g) < Al (27 g) < Fe (56 g) < Ag (108 g)

Takeaway: Equal moles of different elements have different masses. The mass equals the molar mass in grams.

Example 33: Reactions in Solutions — Neutralisation

What volume of 0.1 M $\text{H}_2\text{SO}_4$ is needed to neutralise 50 mL of 0.2 M NaOH?

$$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

Solution:

1. Moles of NaOH = $0.2 \times 0.05$ = 0.01 mol
2. From equation: 1 mol $\text{H}_2\text{SO}_4$ : 2 mol NaOH
3. Moles of $\text{H}_2\text{SO}_4$ = 0.01/2 = 0.005 mol
4. Volume = 0.005/0.1 = 0.05 L = 50 mL

Final Answer: 50 mL of 0.1 M $\text{H}_2\text{SO}_4$