Chapter 1: Some Basic Concepts of Chemistry Application Questions | chemistry Class 11 CBSE

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Chapter 1: Some Basic Concepts of Chemistry Application Questions | chemistry Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A chemist analyses a white powder sample from a crime scene and finds it contains sodium (Na) and chlorine (Cl) in a mass ratio of 23:35.5. Can the chemist conclude this is pure sodium chloride? How does the Law of Definite Proportions help here?

Yes, the chemist can use the Law of Definite Proportions — a given compound always contains the same elements in the same mass ratio — to confirm the identity. Pure NaCl always has Na:Cl = 23:35.5, so matching this ratio strongly suggests it is sodium chloride.
This law implies that if the sample were adulterated with another substance, the observed Na:Cl ratio would deviate from the theoretical value of 23:35.5; even a single elemental ratio mismatch confirms impurity.
The formula mass of NaCl = 23.0 + 35.5 = 58.5 u, so sodium comprises (23/58.5) × 100 = 39.3% and chlorine 60.7% by mass; the chemist can compare measured mass percentages with these theoretical values.
This approach is routinely used in analytical chemistry to verify compound purity by checking whether the elemental composition matches the theoretical value for the pure substance.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A student measures the mass of a sample three times on an analytical balance and gets 9.42 g, 9.43 g, and 9.42 g. The true mass is 9.50 g. Comment on the precision and accuracy of the measurements.

The measurements (9.42, 9.43, 9.42 g) are very close to each other, with a range of only 0.01 g; this means the measurements are highly precise — they show good reproducibility and consistency.
However, the true value is 9.50 g, and all three measurements are significantly lower (by ~0.08 g); this indicates the measurements are not accurate — they do not agree with the true value.
This is a classic example of systematic error (e.g., an uncalibrated balance showing consistently low values) — the instrument is reliable (precise) but not correctly adjusted (not accurate).
To improve accuracy, the student should recalibrate the analytical balance using a standard reference weight; precision is already satisfactory and requires no change in measurement technique.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A factory produces 500 kg of ammonia daily using the reaction N2 + 3H2 → 2NH3. With reference to stoichiometry and the mole concept, calculate the mass of nitrogen and hydrogen consumed daily.

Molar masses: N2 = 28 g mol-1; H2 = 2 g mol-1; NH3 = 17 g mol-1. Moles of NH3 produced = 500,000 g / 17 g mol-1 = 29,412 mol.
From the balanced equation: 2 mol NH3 is produced from 1 mol N2; so moles of N2 required = 29,412 / 2 = 14,706 mol. Mass of N2 = 14,706 × 28 = 411,768 g ≈ 411.8 kg.
3 mol H2 produces 2 mol NH3; so moles of H2 required = (3/2) × 29,412 = 44,118 mol. Mass of H2 = 44,118 × 2 = 88,236 g ≈ 88.2 kg.
These calculations demonstrate the industrial importance of stoichiometry: knowing exact masses of raw materials (N2 and H2) required prevents waste and ensures efficient production in the Haber process.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to concentration expressions, explain what happens to the molarity and molality of a NaCl solution when it is heated from 25°C to 50°C. Which concentration unit would a scientist prefer for temperature-sensitive experiments?

When the NaCl solution is heated, the solution expands — its volume increases. Since molarity = moles of solute / volume of solution (L), and the number of moles of NaCl is unchanged while the volume increases, the molarity decreases.
Molality = moles of solute / mass of solvent (kg); upon heating, the mass of solvent (water) does not change. Therefore, the molality remains constant regardless of temperature change.
For temperature-sensitive experiments — such as colligative property measurements (boiling point elevation, freezing point depression), thermodynamic calculations, or reactions at varying temperatures — molality is the preferred concentration unit because it is temperature-independent.
Molarity is convenient for preparing standard solutions in the lab at room temperature, but for reactions conducted at high temperatures or in environments with significant temperature variation, molality ensures the concentration value remains valid throughout the experiment.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A pharmacist prepares a 0.2 M NaOH solution from a 1 M stock solution. Calculate the volume of stock solution required to make 1 litre of the dilute solution, using the concept of molarity.

Using the dilution formula: M1V1 = M2V2, where M1 = 1.0 M (stock), V1 = volume to be calculated, M2 = 0.2 M (desired), V2 = 1000 mL (1 litre).
Substituting: 1.0 M × V1 = 0.2 M × 1000 mL; V1 = (0.2 × 1000) / 1.0 = 200 mL.
The pharmacist should measure 200 mL of the 1 M NaOH stock solution and add sufficient water to bring the total volume to 1000 mL (1 litre), giving a 0.2 M solution.
The principle underlying this is that the number of moles of NaOH remains constant upon dilution (0.2 mol); only the volume of solvent (water) increases, thereby decreasing the concentration from 1.0 M to 0.2 M.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

A drinking water sample is found to contain chloroform (CHCl3) at 15 ppm by mass. Express this in mass per cent and comment on the significance using the mole concept.

15 ppm (by mass) means 15 parts of CHCl3 per million parts of water sample. Mass per cent = (15 / 1,000,000) × 100 = 1.5 × 10-3 % = 0.0015%.
In 1 kg (1,000,000 mg) of water sample, there are 15 mg of CHCl3. Molar mass of CHCl3 = 12 + 1 + 3(35.5) = 119.5 g mol-1; molality = (0.015 g / 119.5 g mol-1) / 1 kg solvent ≈ 1.255 × 10-4 m.
Although the mass per cent appears tiny (0.0015%), even such low concentrations can be detected using modern analytical tools like mass spectrometry — illustrating why significant figures and scientific notation matter in environmental monitoring.
CHCl3 is considered carcinogenic; this problem demonstrates how the mole concept and concentration calculations have real-world health implications — even trace amounts of harmful substances must be quantified precisely to enforce safety standards.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

Carbon exists as three isotopes: 12C (98.892%, 12 u), 13C (1.108%, 13.00335 u), and 14C (~0%, 14.00317 u). Calculate the average atomic mass of carbon and explain why this value is used in all chemical calculations.

Average atomic mass = (fraction of 12C × mass of 12C) + (fraction of 13C × mass of 13C) + (fraction of 14C × mass of 14C).
= (0.98892 × 12) + (0.01108 × 13.00335) + (2 × 10-12 × 14.00317) = 11.867 + 0.14408 + negligible = 12.011 u.
This value (12.011 u) represents the mass of a 'typical' carbon atom as it occurs in nature — a weighted average of all naturally occurring isotopes. In any laboratory sample of carbon, atoms of all three isotopes are present in these natural proportions.
Using 12.011 u in calculations (rather than 12 u for pure 12C) gives results that reflect actual experimental outcomes — amounts measured in the lab correspond to naturally occurring carbon with all its isotopes, not just the 12C standard.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A metallurgist wants to extract copper from copper sulphate (CuSO4). If 100 g of CuSO4 is available, calculate the maximum mass of copper that can be obtained.

Molar mass of CuSO4: Cu = 63.5, S = 32, O = 64; total = 63.5 + 32 + 64 = 159.5 g mol-1. Moles of CuSO4 = 100 g / 159.5 g mol-1 = 0.627 mol.
Each formula unit of CuSO4 contains exactly one Cu atom; therefore, moles of Cu = 0.627 mol. Mass of Cu = 0.627 mol × 63.5 g mol-1 = 39.8 g.
Mass per cent of Cu in CuSO4 = (63.5/159.5) × 100 = 39.8%; this confirms that at most 39.8 g of copper can be extracted from 100 g of CuSO4.
This calculation illustrates the practical application of stoichiometry and percentage composition in industrial metallurgy — knowing the theoretical yield helps evaluate the efficiency of the extraction process.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

Calcium carbonate (CaCO3) reacts with 25 mL of 0.75 M HCl: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). Calculate the mass of CaCO3 required for complete reaction.

Moles of HCl = Molarity × Volume (L) = 0.75 mol L-1 × 0.025 L = 0.01875 mol HCl.
From the balanced equation, 1 mol CaCO3 reacts with 2 mol HCl; therefore, moles of CaCO3 required = 0.01875 / 2 = 0.009375 mol.
Molar mass of CaCO3 = 40 + 12 + 48 = 100 g mol-1. Mass of CaCO3 = 0.009375 mol × 100 g mol-1 = 0.9375 g ≈ 0.94 g.
This problem demonstrates how molarity links volume of solution to moles of solute, enabling stoichiometric calculations; this type of calculation is fundamental in volumetric analysis (titrations) in the chemistry laboratory.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

With reference to the Law of Conservation of Mass, explain why a campfire seems to lose mass as wood burns but a chemist would argue no mass is actually lost.

Lavoisier's Law of Conservation of Mass (1789) states that in any chemical change, total mass of reactants equals total mass of products; matter is neither created nor destroyed.
In burning wood (combustion), the carbon and hydrogen in wood react with oxygen from the air to produce carbon dioxide (CO2), water vapour (H2O), and ash. The CO2 and H2O are gases that escape into the atmosphere and are invisible.
If the ash remaining, plus all CO2 and H2O produced, plus the oxygen consumed from air are all accounted for and summed, the total mass equals the initial mass of wood plus oxygen consumed.
The apparent 'loss' of mass is an observational illusion caused by not accounting for the gaseous products; this is precisely why Lavoisier used closed systems in his experiments to carefully measure all reactants and products.

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Chapter 1: Some Basic Concepts of Chemistry | Application Questions

A chemist analyses a white powder sample from a crime scene and finds it contains sodium (Na) and chlorine (Cl) in a mass ratio of 23:35.5. Can the chemist conclude this is pure sodium chloride? How does the Law of Definite Proportions help here?

A student measures the mass of a sample three times on an analytical balance and gets 9.42 g, 9.43 g, and 9.42 g. The true mass is 9.50 g. Comment on the precision and accuracy of the measurements.

A factory produces 500 kg of ammonia daily using the reaction N2 + 3H2 → 2NH3. With reference to stoichiometry and the mole concept, calculate the mass of nitrogen and hydrogen consumed daily.

With reference to concentration expressions, explain what happens to the molarity and molality of a NaCl solution when it is heated from 25°C to 50°C. Which concentration unit would a scientist prefer for temperature-sensitive experiments?

A pharmacist prepares a 0.2 M NaOH solution from a 1 M stock solution. Calculate the volume of stock solution required to make 1 litre of the dilute solution, using the concept of molarity.

A drinking water sample is found to contain chloroform (CHCl3) at 15 ppm by mass. Express this in mass per cent and comment on the significance using the mole concept.

Carbon exists as three isotopes: 12C (98.892%, 12 u), 13C (1.108%, 13.00335 u), and 14C (~0%, 14.00317 u). Calculate the average atomic mass of carbon and explain why this value is used in all chemical calculations.

A metallurgist wants to extract copper from copper sulphate (CuSO4). If 100 g of CuSO4 is available, calculate the maximum mass of copper that can be obtained.

Calcium carbonate (CaCO3) reacts with 25 mL of 0.75 M HCl: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). Calculate the mass of CaCO3 required for complete reaction.

With reference to the Law of Conservation of Mass, explain why a campfire seems to lose mass as wood burns but a chemist would argue no mass is actually lost.

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