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Class 10 chemistry chapter 15 stoichiometry model paper pdf | Test 1
Class 10 chemistry chapter 15 stoichiometry model paper solutions
Prepare for the Punjab Board Class 10 Chemistry examination with the Chapter 15: Stoichiometry Class Test 1 from the Punjab Board 2027 Model Papers. This Chemistry test helps students practice important Stoichiometry concepts, numerical problems, chemical equations, and related questions according to the Class 10 Chemistry syllabus. Download or attempt this Class 10 Chemistry Chapter 15 Test 1 to improve your preparation for the Punjab Board annual examination 2027.
1. Room temperature and pressure (RTP) are defined as:
- A. 0 °C and 1 atm
- B. 25 °C and 760 torr
- C. 100 °C and 1 atm
- D. 273 K and 1 atm
Answer: B. RTP = 25 °C (298 K) and 1 atm (760 torr).
2. One mole of any gas occupies a volume at RTP.
- A. 22.4 dm3
- B. 25 dm3
- C. 24 dm3
- D. 1 dm3
Answer: C. 1 mole of any gas = 24 dm3 at RTP (22.4 dm3 is at STP).
3. If 20 g of NaCl is dissolved in 0.1 dm3 of water, the concentration is:
- A. 10 g/dm3
- B. 40 g/dm3
- C. 100 g/dm3
- D. 200 g/dm3
Answer: D. Concentration = 20 g ÷ 0.1 dm3 = 200 g/dm3.
4. What is the volume of NaOH used in dm3 if it is 15 cm3?
- A. 0.15 dm3
- B. 0.015 dm3
- C. 0.0015 dm3
- D. 1.5 dm3
Answer: B. 15 cm3 ÷ 1000 = 0.015 dm3.
5. The formula which shows the simplest whole number ratio of atoms in a compound is called:
- A. Molecular formula
- B. Structural formula
- C. Empirical formula
- D. Chemical equation
Answer: C. Empirical formula.
6. The molecular formula of sulphur is:
- A. S
- B. S2
- C. S4
- D. S8
Answer: D. Sulphur exists as S8 molecules (crown-shaped ring).
7. The molecular formula of benzene is:
- A. CH
- B. C2H2
- C. C6H6
- D. C6H12
Answer: C. Benzene = C6H6 (empirical formula CH).
8. The mass of 0.848 mol of CO2 is approximately:
- A. 18 g
- B. 36.96 g
- C. 44 g
- D. 73 g
Answer: B. Mass = 0.848 mol × 44 g/mol ≈ 37 g, so the closest option is 36.96 g.
786Times • Chemistry Chapter-15 • Class Test-1 (Objective)
Attempt any 10 Questions.10×2=20
i. What is meant by room temperature and pressure (RTP)?
Ans: RTP means a temperature of 25 °C (298 K) and a pressure of 1 atm (760 torr / 101.3 kPa). At RTP, 1 mole of any gas occupies 24 dm3.
ii. Calculate the volume of 1.5 mole of hydrogen at RTP.
Ans: 1 mole of H2 = 24 dm3
Volume = 1.5 × 24 = 36 dm3
Volume = 1.5 × 24 = 36 dm3
iii. Calculate the concentration if 30 g of solute is dissolved in 0.2 dm3 of solution.
Ans: Concentration = mass ÷ volume = 30 ÷ 0.2 = 150 g/dm3
iv. Find the number of moles of 15 cm3 of solution of NaOH having molar concentration is 0.5 dm3.
Ans: (Molarity taken as 0.5 mol/dm3)
Volume = 15 cm3 = 0.015 dm3
Moles = M × V = 0.5 × 0.015 = 0.0075 mol
Volume = 15 cm3 = 0.015 dm3
Moles = M × V = 0.5 × 0.015 = 0.0075 mol
v. Find percentage composition of Hydrogen if 8.657 gram of compound decompose to give 1.962 gram of hydrogen.
Ans: % of H = (1.962 ÷ 8.657) × 100 = 22.66 %
vi. What is an empirical formula? Give an example.
Ans: The formula that shows the simplest whole number ratio of atoms in a compound. Example: the empirical formula of glucose C6H12O6 is CH2O.
vii. How is molecular formula related to empirical formula?
Ans: Molecular formula = (Empirical formula)n, where n = molecular mass ÷ empirical formula mass. Example: n = 6 for glucose (CH2O)6 = C6H12O6.
viii. Write the steps to identify a limiting reactant.
Ans:
- Write the balanced chemical equation.
- Convert the given masses of reactants into moles.
- Divide each number of moles by its coefficient in the balanced equation.
- The reactant giving the smallest value is the limiting reactant.
ix. What is theoretical yield?
Ans: The maximum amount of product calculated from the balanced equation, assuming the reaction goes to completion with no loss.
x. Define percentage yield. Write it’s formula.
Ans: The ratio of the actual yield to the theoretical yield, multiplied by 100.
% Yield = (Actual yield ÷ Theoretical yield) × 100
% Yield = (Actual yield ÷ Theoretical yield) × 100
SECTION-II (4+4)
1.
An unknown hydrocarbon contains 85.71% carbon. Its molar mass is 84 g/mol. What is its molecular formula?(Quick check-15.6)
Ans: C = 85.71 %, H = 100 − 85.71 = 14.29 %
Moles: C = 85.71 ÷ 12 = 7.14; H = 14.29 ÷ 1 = 14.29
Ratio (÷ 7.14): C : H = 1 : 2 → empirical formula = CH2 (mass 14)
n = 84 ÷ 14 = 6
Molecular formula = (CH2)6 = C6H12
Moles: C = 85.71 ÷ 12 = 7.14; H = 14.29 ÷ 1 = 14.29
Ratio (÷ 7.14): C : H = 1 : 2 → empirical formula = CH2 (mass 14)
n = 84 ÷ 14 = 6
Molecular formula = (CH2)6 = C6H12
2.
Calculate the number of moles, volume and number of molecules in 4.0 g of CH4 at RTP.(Num 15.1)
Ans: Molar mass of CH4 = 12 + 4 = 16 g/mol
Moles = 4.0 ÷ 16 = 0.25 mol
Volume = 0.25 × 24 = 6 dm3
Molecules = 0.25 × 6.022 × 1023 = 1.5 × 1023
Moles = 4.0 ÷ 16 = 0.25 mol
Volume = 0.25 × 24 = 6 dm3
Molecules = 0.25 × 6.022 × 1023 = 1.5 × 1023
786Times • Chemistry Chapter-15 • Class Test-1 (Subjective)
