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Class 10 chemistry chapter 15 stoichiometry pdf free download | Test 2
Class 10 Chemistry Chapter 15 Stoichiometry Solutions
Prepare for the Punjab Board Class 10 Chemistry annual examination with the Chapter 15: Stoichiometry Class Test 2 from the Punjab Board 2027 Model Papers. This test helps students practice important Stoichiometry concepts, chemical equations, numerical problems, and exam-related questions according to the Class 10 Chemistry syllabus. Use this Class 10 Chemistry Chapter 15 Test 2 to assess your understanding, strengthen your problem-solving skills, and improve your preparation for the 2027 board examination.
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TIME: 10 MIN
TOTAL MARKS:30
MARKS: 8
1. The value of room temperature in kelvin is:
A. 273 K
B. 298.15 K
C. 300 K
D. 250 K
Answer: B. 298.15 K
2. The amount of solute is usually expressed in:
A. dm3
B. cm3
C. grams or moles
D. litres only
Answer: C. grams or moles
3. Which solution is more concentrated:
A. 0.5M
B. 1M
C. 1.5M
D. 2M
Answer: D. 2M
4. The percentage of calcium in Ca(OH)2 is approximately:
A. 43.24 %
B. 2.70 %
C. 54.05 %
D. 60.26 %
Answer: C. 54.05 % (40 ÷ 74 × 100)
5. The empirical formula of hydrogen peroxide (H2O2) is:
A. H2O
B. H2O2
C. HO
D. H2
Answer: C. HO
6. The empirical formula of benzene is CH and its molecular mass is 78. The value of *n* is:
A. 3
B. 6
C. 13
D. 78
Answer: B. 6 (78 ÷ 13 = 6)
7. Number of moles of glucose in 25 g of glucose is:
A. 1.25 mol
B. 0.138 mol
C. 0.878 mol
D. 6 mol
Answer: B. 0.138 mol (25 ÷ 180)
8. The actual yield of a reaction is always the theoretical yield.
A. Equal to
B. Greater than
C. Less than
D. Double
Answer: C. Less than
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TIME: 22 MIN
MARKS: 22
Attempt any 10 Questions.
10×2=20
i. Define molar volume.
The volume occupied by one mole of any gas at STP (0 °C and 1 atm) is called molar volume. Its value is 22.414 dm3.
ii. What is meant by concentration of a solution?
The amount of solute dissolved in a given quantity of solution (or solvent) is called its concentration, e.g. mol/dm3 (molarity).
iii. What is titration?
A laboratory technique used to find the unknown concentration of a solution by reacting it with a solution of known concentration (standard solution), using an indicator.
iv. Define percentage of an element in a compound.
The mass of an element present in 100 parts by mass of a compound. % of element = (mass of element ÷ molar mass of compound) × 100
v. Why stoichiometry important? Give example.
It tells the exact quantities of reactants and products in a chemical reaction. Example: 2H2 + O2 → 2H2O; 4 g H2 reacts with 32 g O2 to give 36 g H2O.
vi. What is a molecular formula? Give example.
The formula showing the actual number of atoms of each element in one molecule of a compound. Example: glucose C6H12O6, benzene C6H6.
vii. What is meant by a limiting reactant?
The reactant that is completely consumed first in a reaction and limits the amount of product formed.
viii. What is actual yield?
The quantity of product actually obtained in a chemical reaction in the laboratory.
ix. Why is actual yield always less than theoretical yield?
Because of reversible reactions, side reactions, loss of product during filtration/transfer, and impurities in the reactants.
x. Which information given percentage purity? Write it's formula.
The mass of pure substance present in an impure sample. % purity = (mass of pure substance ÷ mass of impure sample) × 100
SECTION-II (4+4)
1. Benzene is represented by its empirical formula CH while its molecular mass is 78. Find out its molecular formula.(Exm. 15.13)
Empirical formula mass of CH = 12 + 1 = 13
n = molecular mass ÷ empirical formula mass = 78 ÷ 13 = 6
Molecular formula = (CH)6 = C6H6
n = molecular mass ÷ empirical formula mass = 78 ÷ 13 = 6
Molecular formula = (CH)6 = C6H6
2. What mass of NaCl need to be dissolved in 150 cm3 of a solution if the concentration of the solution is 0.4 mol/dm3?(Num 15.2)
Volume = 150 cm3 = 0.15 dm3
Moles of NaCl = 0.4 × 0.15 = 0.06 mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = 0.06 × 58.5 = 3.51 g
Moles of NaCl = 0.4 × 0.15 = 0.06 mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = 0.06 × 58.5 = 3.51 g
