CBSE Class 11 Chemistry Some Basic Concepts Worksheet Set 05

Worksheet: Some Basic Concepts of Chemistry

Question 1. Briefly explain the difference between precision and accuracy with an example.
Answer:
Precision refers to the closeness of various measurements of the same quantity to one another.
Accuracy is the agreement of a particular measured value with the true value of the result.
Example: the true value of a length is 2.00 cm. Student A reports 1.95, 1.93 and 1.96 cm: these agree closely (precise) but are far from 2.00 cm (not accurate). Student B reports 1.99, 2.01 and 2.00 cm: these are both precise and accurate.
In simple words: Precision is about repeatability; accuracy is about being close to the true value.

Teacher's Note:
1. A set of readings can be precise without being accurate.
2. Give a numerical example in the board answer.

Question 2. State the number of significant figures in each of the following: (i) 208.91 (ii) 0.00456 (iii) 453 (iv) 0.346 (v) 16.00
Answer:
(i) 208.91 has five significant figures (the zero between non-zero digits counts).
(ii) 0.00456 has three (zeros before the first non-zero digit do not count).
(iii) 453 has three.
(iv) 0.346 has three.
(v) 16.00 has four (zeros to the right of the decimal point after a non-zero digit count).
In simple words: Count from the first non-zero digit; zeros in between and trailing zeros after a decimal point count.

Teacher's Note:
1. Learn the four rules for zeros.
2. Writing the number in scientific notation makes counting easy.

Question 3. Express the following in scientific notation with 2 significant figures: (a) 0.0048 (b) 234,000 (c) 200.0
Answer:
(a) 0.0048 = 4.8 × 10⁻³
(b) 234,000 = 2.34 × 10⁵, which to two significant figures is 2.3 × 10⁵
(c) 200.0 = 2.000 × 10², which to two significant figures is 2.0 × 10²
In simple words: Write the number as N × 10ⁿ with N between 1 and 10, then keep only the required digits.

Teacher's Note:
1. Move the decimal point until one non-zero digit is before it.
2. Moving left makes n positive; moving right makes n negative.

Question 4. What is the unit factor (factor label) method? Express 93 million miles (the distance between the earth and the sun) in metres. (1 mile = 1.60934 km)
Answer:
The unit factor method (dimensional analysis) converts a quantity from one system of units to another by multiplying it by a ratio equal to 1, such as 1.60934 × 10³ m ÷ 1 mile.
93 million miles = 93 × 10⁶ miles.
93 × 10⁶ miles × (1.60934 × 10³ m ÷ 1 mile) = 1.4967 × 10¹¹ m ≈ 1.5 × 10¹¹ m.
In simple words: Multiply by a conversion ratio so that the old unit cancels and the new unit remains.

Teacher's Note:
1. Keep the units in every step so that they cancel properly.
2. Give the final answer to the same number of significant figures as the data (here two).

Teacher's Note:
1. Make a two-column table in the exam.
2. Say that the composition of a compound is fixed but that of a mixture is variable.

Question 6. Carbon and oxygen form two compounds. The carbon content of one is 42.9% and that of the other is 27.3%. Show that this data agree with the law of multiple proportions.
Answer:
Oxide I: carbon = 42.9% and oxygen = 57.1%. Mass of oxygen per 1 g of carbon = 57.1 ÷ 42.9 = 1.33 g.
Oxide II: carbon = 27.3% and oxygen = 72.7%. Mass of oxygen per 1 g of carbon = 72.7 ÷ 27.3 = 2.66 g.
Ratio of the masses of oxygen combining with a fixed mass of carbon = 1.33 : 2.66 = 1 : 2, a simple whole-number ratio.
Hence the law of multiple proportions is verified (the compounds are CO and CO₂).
In simple words: Find the oxygen per gram of carbon in each oxide; the ratio of the two values is simple.

Teacher's Note:
1. Fix the mass of one element at 1 g.
2. A simple ratio such as 1 : 2 or 2 : 3 confirms the law.

Question 7. 2.4 g of iron oxide on reduction with hydrogen gave 1.68 g of iron. In another experiment, 2.9 g of iron oxide gave 2.03 g of iron. Show that the data illustrate the law of constant proportions.
Answer:
First experiment: mass of oxygen = 2.4 − 1.68 = 0.72 g, so oxygen : iron = 0.72 : 1.68 = 1 : 2.33.
Second experiment: mass of oxygen = 2.9 − 2.03 = 0.87 g, so oxygen : iron = 0.87 : 2.03 = 1 : 2.33.
The ratio is the same in both samples, so the data illustrate the law of constant (definite) proportions.
In simple words: Same compound, same mass ratio of its elements, whatever the sample size.

Teacher's Note:
1. Mass of oxygen = mass of oxide − mass of iron.
2. Divide both masses by the smaller one to compare the ratios.

Question 8. Chlorine has two isotopes of atomic masses 34.97 u and 36.97 u. Their relative abundances are 0.755 and 0.245 respectively. Calculate the average atomic mass of chlorine.
Answer:
Average atomic mass = (34.97 × 0.755) + (36.97 × 0.245)
= 26.40 + 9.06 = 35.46 u
In simple words: The average atomic mass is the abundance-weighted mean of the isotopic masses.

Teacher's Note:
1. Multiply each isotopic mass by its fractional abundance and add.
2. The abundances must add up to 1 (or 100%).

Question 9. Calculate the mass per cent of carbon, hydrogen and oxygen in ethanol (C₂H₅OH). (C = 12.01, H = 1.008, O = 16.00)
Answer:
Molar mass of C₂H₅OH = 2(12.01) + 6(1.008) + 16.00 = 46.068 g mol⁻¹.
Mass % of carbon = (24.02 ÷ 46.068) × 100 = 52.14%
Mass % of hydrogen = (6.048 ÷ 46.068) × 100 = 13.13%
Mass % of oxygen = (16.00 ÷ 46.068) × 100 = 34.73%
Check: 52.14 + 13.13 + 34.73 = 100.00.
In simple words: Divide the mass of each element in one mole by the molar mass and multiply by 100.

Teacher's Note:
1. Ethanol has six hydrogen atoms in all (C₂H₅OH).
2. Always check that the percentages add up to 100.

Question 10. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer:
Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol⁻¹.
0.50 mol Na₂CO₃ means an amount of 0.50 × 106 = 53 g of the substance.
0.50 M Na₂CO₃ means a solution containing 0.50 mol (53 g) of Na₂CO₃ in every 1 L of the solution.
In simple words: Mol is an amount; M (molar) is a concentration, amount per litre of solution.

Teacher's Note:
1. Molarity always has volume of solution in its definition.
2. Unit of molarity: mol L⁻¹.

Question 11. Calculate the molarity of NaOH in the solution made by dissolving 4 g of NaOH in enough water to form 250 mL of the solution.
Answer:
Molar mass of NaOH = 40 g mol⁻¹, so moles of NaOH = 4 ÷ 40 = 0.1 mol.
Volume of solution = 250 mL = 0.250 L.
Molarity = 0.1 mol ÷ 0.250 L = 0.4 mol L⁻¹ (0.4 M)
In simple words: Molarity = moles of solute ÷ litres of solution.

Teacher's Note:
1. Convert mL to L before dividing.
2. Write the unit mol L⁻¹ or M in the answer.

Question 12. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g mol⁻¹. What are its empirical and molecular formulae?
Answer:
Taking 100 g: H = 4.07 g, C = 24.27 g, Cl = 71.65 g.
Moles: H = 4.07 ÷ 1.008 = 4.04; C = 24.27 ÷ 12.01 = 2.02; Cl = 71.65 ÷ 35.5 = 2.02.
Dividing by 2.02 gives the ratio H : C : Cl = 2 : 1 : 1, so the empirical formula is CH₂Cl.
Empirical formula mass = 12.01 + 2(1.008) + 35.5 = 49.5 g mol⁻¹. n = 98.96 ÷ 49.5 = 2.
Molecular formula = 2 × CH₂Cl = C₂H₄Cl₂.
In simple words: Convert percentages to moles, divide by the smallest, then scale up using the molar mass.

Teacher's Note:
1. The empirical formula is the simplest whole-number ratio.
2. n = molar mass ÷ empirical formula mass.

Question 13. Dinitrogen and dihydrogen react to form ammonia: N₂(g) + 3H₂(g) → 2NH₃(g). (i) Calculate the mass of ammonia produced if 2.00 × 10³ g of N₂ reacts with 1.00 × 10³ g of H₂. (ii) Will any reactant remain unreacted? If so, which one and how much?
Answer:
(i) Moles of N₂ = 2000 ÷ 28 = 71.4 mol; moles of H₂ = 1000 ÷ 2 = 500 mol.
71.4 mol of N₂ needs 3 × 71.4 = 214.3 mol of H₂, which is less than 500 mol, so N₂ is the limiting reagent.
Moles of NH₃ formed = 2 × 71.4 = 142.9 mol, i.e. 142.9 × 17 = 2.43 × 10³ g (2428.6 g).
(ii) Yes, H₂ is in excess. H₂ used = 214.3 mol = 428.6 g, so H₂ left over = 1000 − 428.6 = 571.4 g.
In simple words: The reactant that finishes first (N₂ here) decides how much ammonia forms.

Teacher's Note:
1. Check the mole ratio 1 : 3 to find the limiting reagent.
2. Mass left over = mass taken − mass used.

Question 14. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g of carbon dioxide and 0.690 g of water and no other products. 10.0 L of the gas (at STP) weighs 11.6 g. Calculate (i) the empirical formula, (ii) the molar mass and (iii) the molecular formula of the gas.
Answer:
(i) Carbon in 3.38 g of CO₂ = 3.38 × 12 ÷ 44 = 0.9218 g. Hydrogen in 0.690 g of H₂O = 0.690 × 2 ÷ 18 = 0.0767 g.
Moles: C = 0.9218 ÷ 12 = 0.0768; H = 0.0767 ÷ 1 = 0.0767. The ratio C : H = 1 : 1, so the empirical formula is CH.
(ii) Mass of 22.4 L of the gas = 11.6 × 22.4 ÷ 10.0 = 25.98 ≈ 26 g, so the molar mass is 26 g mol⁻¹.
(iii) Empirical formula mass of CH = 13, so n = 26 ÷ 13 = 2. Molecular formula = C₂H₂.
In simple words: Find carbon from CO₂ and hydrogen from H₂O, then get the formula from the mole ratio and the molar mass.

Teacher's Note:
1. All the carbon of CO₂ and all the hydrogen of H₂O came from the fuel.
2. Use 22.4 L mol⁻¹ at STP to get the molar mass of a gas.

Question 15. Calculate the number of atoms in each of the following: (i) 52 mol of Ar (ii) 52 u of He (iii) 52 g of He
Answer:
(i) 1 mol of Ar has 6.022 × 10²³ atoms, so 52 mol has 52 × 6.022 × 10²³ = 3.131 × 10²⁵ atoms.
(ii) One atom of He has a mass of 4 u, so 52 u of He = 52 ÷ 4 = 13 atoms.
(iii) Molar mass of He = 4 g mol⁻¹, so 52 g = 13 mol, and the number of atoms = 13 × 6.022 × 10²³ = 7.83 × 10²⁴ atoms.
In simple words: Moles times Avogadro's number gives atoms; mass in u divided by the atomic mass in u also gives atoms.

Teacher's Note:
1. Do not mix up u (mass of one atom) and g (mass of one mole of atoms).
2. Helium is monoatomic.