Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry worksheet with answer key
A 20-mark worksheet on Some Basic Concepts of Chemistry for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) 20.60 g
In 0.02060 the leading zeros are not significant, the zero between 2 and 6 is, and the terminal zero after the decimal point is: four figures (2, 0, 6, 0). 20.60 also has four. 2060 (no decimal point) has three, and 2.06 and 0.206 have three.
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Answer: (c) whether it burns in air to form magnesium oxide
Combustibility is a chemical property: observing it needs the magnesium to change into a new substance, magnesium oxide. Colour, density and melting point are physical properties, measured without changing the identity of the metal.
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Answer: (c) −40°
Put °F = °C = x in °F = (9/5)(°C) + 32: x = 9x/5 + 32, so −4x/5 = 32 and x = −40. So −40 °C = −40 °F.
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Answer: (a) 450 pm
1 nm = 10⁻⁹ m and 1 pm = 10⁻¹² m, so 0.45 nm × (10⁻⁹ m / 1 nm) × (1 pm / 10⁻¹² m) = 0.45 × 10³ pm = 450 pm.
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Answer: (d) 16.4 g
The sum is 16.423 g. In addition the result may carry no more decimal places than the least precise figure, 4.2 g (one decimal place), so it is reported as 16.4 g.
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(i) (3.0 × 4.0) × 10⁻⁴⁺⁷ = 12 × 10³ = 1.2 × 10⁴. (ii) Write both with the same exponent: 7.2 × 10⁵ − 0.60 × 10⁵ = (7.2 − 0.60) × 10⁵ = 6.6 × 10⁵.
- (i) multiply digit terms, add exponents: 1.2 × 10⁴ (1)
- (ii) same exponent first, then subtract: 6.6 × 10⁵ (1)
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By M₁V₁ = M₂V₂: 2.0 M × 25 mL = 0.50 M × V₂, so V₂ = 100 mL. The moles of HCl stay the same, so water to be added = 100 − 25 = 75 mL.
- M₁V₁ = M₂V₂ gives final volume 100 mL (1)
- water added = 100 − 25 = 75 mL (1)
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Both have C : H = 1 : 1 as the simplest whole-number ratio, so the empirical formula is CH. The molecular formula gives the exact number of each kind of atom in one molecule: two C and two H in ethyne, six C and six H in benzene.
- empirical formula CH (simplest whole-number ratio 1 : 1) (1)
- molecular formula gives the exact number of each kind of atom in a molecule (1)
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(i) Chemical reactions involve reorganisation of atoms, which are neither created nor destroyed in a reaction, so total mass does not change. (ii) Compounds are formed when atoms of different elements combine in a fixed ratio, and all atoms of an element have identical mass, so a compound has a fixed composition by mass. The theory could not explain the law of gaseous volumes (nor why atoms combine).
- (i) atoms rearranged, neither created nor destroyed (1)
- (ii) atoms of identical mass combine in a fixed ratio (1)
- could not explain gaseous volumes (or the reason atoms combine) (1)
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(i) Mg = 6.0/24 = 0.25 mol; O₂ = 6.0/32 = 0.1875 mol. 0.25 mol Mg needs only 0.125 mol O₂, so magnesium is used up first: Mg is the limiting reagent. (ii) MgO formed = 0.25 mol × 40 g mol⁻¹ = 10.0 g. (iii) O₂ left = 0.1875 − 0.125 = 0.0625 mol, which needs 2 × 0.0625 = 0.125 mol Mg = 3.0 g.
- moles compared: Mg (0.25 mol) is limiting (1)
- MgO = 10.0 g (1)
- O₂ left 0.0625 mol needs 3.0 g Mg (1)
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(i) With the same 1.24 g of phosphorus, the masses of oxygen are 0.96 g and 1.60 g, in the ratio 0.96 : 1.60 = 0.6 = 3 : 5, a ratio of small whole numbers: the law of multiple proportions. (ii) P = 1.24/31 = 0.040 mol. In A, O = 0.96/16 = 0.060 mol, so P : O = 0.040 : 0.060 = 2 : 3 and the empirical formula is P₂O₃ (empirical formula mass 110). In B, O = 1.60/16 = 0.100 mol, so P : O = 2 : 5, P₂O₅ (142). (iii) A: n = 220/110 = 2, molecular formula P₄O₆. B: n = 284/142 = 2, molecular formula P₄O₁₀.
- (i) oxygen masses for fixed phosphorus in ratio 3 : 5 (1)
- (ii) empirical formulas P₂O₃ and P₂O₅ (1)
- (iii) n = 2 for both: P₄O₆ and P₄O₁₀ (1)