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Class 10 Chemistry Chapter 15 Stoichiometry Test 5 – Board Pattern 2027
Grade 10 Chemistry Chapter 15 Solved 15.1 to 15.4
Prepare for the Punjab Board Class 10 Chemistry examination with Chapter 15: Stoichiometry Class Test 5 from the Punjab Board 2027 Model Papers. This test covers topics 15.1 to 15.4 and helps students revise important Stoichiometry concepts, chemical equations, mole calculations, and numerical problems included in their syllabus. Practice this Class 10 Chemistry Chapter 15 Test 5 to strengthen your conceptual understanding and improve your preparation for the Punjab Board Annual Examination 2027.
Chapter Wise Class Test
T-5
Chapter-15
Stoichiometry
1.1 Choose the correct Answer.
(Time: 7 Minutes)
(Total Marks: 8)
1.Room Temperature and Pressure (RTP) refers to:
(a) 0°C and 1 atm
(b) 25°C and 760 torr
(c) 100°C and 1 atm
(d) 298 K and 2 atm
2.Which of the following quantity remains the same for one mole of all gases at RTP?
(a) Mass
(b) Size of molecules
(c) Volume
(d) Density
3.In which units is the amount of solute commonly expressed?
(a) Liters only
(b) Grams or moles
(c) Kelvin
(d) Pascal
4.Which change will increase the concentration of a solution?
(a) Adding more solvent
(b) Removing some solute
(c) Adding more solute
(d) Increasing volume only
5.What is the volume of NaOH used in dm3 if it is 15cm3 ?
(a) 0.15 dm3
(b) 0.015 dm3
(c) 0.0015 dm3
(d) 1.5 dm3
15 cm3 ÷ 1000 = 0.015 dm3.
6.Percentage composition is useful in determining:
(a) only molar mass
(b) only density
(c) empirical and molecular formulae
(d) volume of gases
7.The formula mass of Ca(OH)2 is:
(a) 56
(b) 72
(c) 74
(d) 76
40 + 2(16 + 1) = 74 amu.
8.If a compound contains 60 g of carbon in 100 g of compound, its percentage of carbon is:
(a) 40%
(b) 50%
(c) 60%
(d) 90%
(60 ÷ 100) × 100 = 60%.
Chapter Wise Class Test
T-5
Chapter-15
Stoichiometry
Total Marks: 17
(Subjective Type)
Time: 33 Minutes
2.Write short answers to any six (6) questions.[6 × 2 = 12]
(i)
Define Room Temperature and Pressure (RTP).
RTP is the standard condition of 25°C (298 K) temperature and 1 atm (760 torr, 101.3 kPa) pressure, at which one mole of any gas occupies 24 dm3.
(ii)
How does the concept of molar volume help in chemical calculations?
One mole of any gas occupies the same volume (24 dm3 at RTP), so gas volumes can be converted directly into moles, mass or number of particles and compared with the mole ratio of a balanced equation.
(iii)
Calculate the number of moles of nitrogen gas present when it occupies a volume of 2.5 dm3 at RTP.
Moles = 2.5 ÷ 24 = 0.104 mol.
(iv)
What is concentration of a solution? Write its units.
Concentration is the amount of solute dissolved in a given volume of solution. Units: g/dm3 or mol/dm3.
(v)
A solution is prepared by dissolving 10 g of NaOH in 1.2 dm3 of water. Calculate the concentration of the solution in mol/dm3.
Moles NaOH = 10 ÷ 40 = 0.25 mol. Concentration = 0.25 ÷ 1.2 = 0.208 mol/dm3.
(vi)
What information is required to calculate concentration by titration?
The balanced chemical equation (mole ratio), the volume and concentration of the known solution, and the volume of the other solution used at the end point.
(vii)
Calculate the formula mass of Ca(OH)2.
40 + 2(16 + 1) = 74 amu.
(viii)
Calculate the percentage of sulphate ions (SO42−) in Al2(SO4)3.
Formula mass = 2(27) + 3(96) = 342. Sulphate = 288. % = (288 ÷ 342) × 100 = 84.2%.
(ix)
Calculate concentration of a solution in g/dm3 which contains 20 g of sodium chloride dissolved in 0.2 dm3 of water.
20 ÷ 0.2 = 100 g/dm3.
Note: Attempt any one question.[5 × 1 = 5]
3.
Calculate the volume occupied by 200 g carbon dioxide at RTP.
Molar mass CO2 = 44 g/mol. Moles = 200 ÷ 44 = 4.545 mol. Volume = 4.545 × 24 = 109.1 dm3.
4.
What is percentage composition. How is it calculated?
Percentage composition is the percentage by mass of each element in a compound.
% of element = (mass of element in 1 mole ÷ molar mass of compound) × 100
Example: H2O → H = 11.19%, O = 88.81%.
% of element = (mass of element in 1 mole ÷ molar mass of compound) × 100
Example: H2O → H = 11.19%, O = 88.81%.
