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CBSE · Class 11 · Chemistry

Some Basic Concepts of Chemistry

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
One mole of a substance contains:
  1. 22.7 particles
  2. 1 gram of particles
  3. 6.022 × 1022 particles
  4. 6.022 × 1023 particles
2.
How many significant figures are there in 90007?
  1. 5
  2. 6
  3. 4
  4. 7
3.
How many moles are there in 18.5 g of Ca(OH)2? (Atomic masses: H = 1 u, O = 16 u, Ca = 40 u)
  1. 0.125 mol
  2. 0.5 mol
  3. 1 mol
  4. 0.25 mol
4.
How many atoms in total are there in 4 mol of CH4? (N_A = 6.022 × 1023 mol−1)
  1. 1.204 × 1025
  2. 2.409 × 1024
  3. 1.445 × 1025
  4. 4.818 × 1024
5.
18 g of glucose (C6H12O6, molar mass 180 g/mol) is dissolved in water to make 250 mL of solution. Its molarity is:
  1. 0.4 M
  2. 0.8 M
  3. 4 M
  4. 0.04 M
6.
The percentage by mass of O in Na2CO3 is: (Atomic masses: C = 12 u, O = 16 u, Na = 23 u)
  1. 54.72%
  2. 15.09%
  3. 43.40%
  4. 45.28%
7.
The empirical formula of C6H6 is:
  1. CH
  2. C3H3
  3. C6H6
  4. C2H2
8.
Assertion (A): 44 g of CO2 and 30 g of C2H6 contain the same number of molecules. (Atomic masses: H = 1 u, C = 12 u, O = 16 u)
Reason (R): Equal numbers of moles of any substances contain equal numbers of molecules.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
State the law of multiple proportions with an example.
10.
How many moles and how many atoms are there in 310.5 g of lead? (Pb = 207 u; N_A = 6.022 × 1023 mol−1)
11.
27.38 g of hydrogen chloride (HCl, molar mass 36.5 g/mol) is dissolved in 500 g of water. Find the molality of the solution.
12.
230 g of ethanol (C2H5OH) is mixed with 90 g of water. Find the mole fraction of each. (Molar masses: C2H5OH 46 g/mol, water 18 g/mol)
13.
A compound contains P 43.66%, O 56.34% by mass. Find its empirical formula. (Atomic masses: O = 16 u, P = 31 u)
14.
What volume of water must be added to 20 mL of 2 M nitric acid (HNO3) to dilute it to 0.1 M?
15.
What mass of copper sulphate (CuSO4, molar mass 159.5 g/mol) is needed to make 50 mL of a 1 M solution?
16.
A compound has the empirical formula CH2 and a molar mass of 84 g/mol. Find its molecular formula. (Atomic masses: H = 1 u, C = 12 u)
Section C — Word problems4 × 4 = 16 marks
17.
State the laws of chemical combination. Explain the law of definite proportions with an example.
18.
56 g of Fe reacts with 64 g of S by Fe + S → FeS. (a) Which is the limiting reagent? (b) What mass of FeS forms? (c) What mass of the other reactant is left over? (Atomic masses: S = 32 u, Fe = 56 u)
19.
58 g of butane (C4H10) burns completely in oxygen. (a) Write the balanced equation. (b) Find the masses of carbon dioxide and water formed. (c) What mass of oxygen is used? (Atomic masses: H = 1 u, C = 12 u, O = 16 u)
20.
A solution of silver nitrate (AgNO3, molar mass 170 g/mol) is 30% by mass and has a density of 1.4 g/mL. Find (a) its molarity, (b) its molality. (2 decimal places where needed.)
AmbifulWorksheets
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CBSE · Class 11 · Chemistry

Answer key

Some Basic Concepts of Chemistry · AW-11C-SBC-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 6.022 × 1023 particlesthe Avogadro constant
2(a) 5non-zero digits and zeros between them count; leading zeros never count; trailing zeros count after a decimal point
3(d) 0.25 moln = mass ÷ molar mass = 18.5 ÷ 74
4(a) 1.204 × 1025molecules = 4 × N_A; each CH4 has 5 atoms
5(a) 0.4 Mmoles = 18 ÷ 180 = 0.1; ÷ 0.25 L
6(d) 45.28%3 × 16 ÷ 106 × 100
7(a) CHdivide every subscript by their highest common factor
8(a) Both A and R are true, and R is the correct explanation of A.A is true: both are 1 mol. R is true and explains A.
9When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. Example: 12 g of carbon combines with 16 g of oxygen in CO and 32 g in CO2, a ratio of 1 : 2.—
101.5 mol; 9.033 × 1023 atomsn = mass ÷ molar mass; atoms = n × N_A
111.5 mol/kgmoles = 0.75; molality = moles ÷ 0.5 kg
12x(ethanol) = 0.5; x(water) = 0.5moles: 5 and 5; total 10
13P2O5divide each percentage by the atomic mass, then by the smallest result, and round to whole numbers
14380 mL (final volume 400 mL)M1V1 = M2V2
157.975 gmoles = 1 × 0.05
16C6H12n = 84 ÷ 14 = 6
17Laws: conservation of mass; definite proportions; multiple proportions; Gay-Lussac’s law of gaseous volumes; Avogadro’s law. Definite proportions: a compound always contains the same elements combined in the same proportion by mass, whatever its source. Example: pure water from any source has hydrogen and oxygen in the ratio 1 : 8 by mass.—
18(a) Fe; (b) 88 g; (c) 32 g of Smoles ÷ coefficient: Fe 1, S 2; the smaller is limiting
19(a) 2C4H10 + 13O2 → 8CO2 + 10H2O; (b) 176 g CO2, 90 g H2O; (c) 208 g O21 mol of C4H10 reacts; use the mole ratio 2 : 13 : 8 : 10
20(a) 2.47 M; (b) 2.52 mol/kg100 g of solution has 30 g solute (0.18 mol) and 70 g water; its volume is 100 ÷ 1.4 mL