Class 10 Chemistry Chapter 15 Stoichiometry Test 3 – Free download
Grade 10 Chemistry Chapter 15 Model Papers Solutions
Prepare for the Punjab Board Class 10 Chemistry examination with the Chapter 15: Stoichiometry Class Test 3 from the Punjab Board 2027 Model Papers. This Class 10 Chemistry test helps students revise important Stoichiometry concepts, chemical equations, numerical problems, and related topics from the new syllabus. Practice the Chemistry Chapter 15 Test 3 to strengthen your subject knowledge and improve your preparation for the Punjab Board Class 10 annual examination 2027.
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1.
Volume occupied by 15 moles of ammonia gas at RTP:(A) 224 dm³ (B) 160 dm³ (C) 360 dm³ (D) 265 dm³Ans: (C) 360 dm³ — 15 mol × 24 dm³/mol = 360 dm³
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2.
The volume of a gas varies with changes in:(A) Colour and odour (B) Pressure and temperature (C) Mass and size of molecules (D) Number of protonsAns: (B) Pressure and temperature
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3.
How many moles of natural gas CH4 will be present in a cylinder if its volume is 18 dm³ at RTP?(A) 0.75 mol (B) 0.67 mol (C) 0.85 mol (D) 0.56 molAns: (A) 0.75 mol — moles = 18 ÷ 24 = 0.75 mol
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4.
The percentage of an element in a compound is the number of grams of that element present in ________ of the compound.(A) 10 grams (B) 50 grams (C) 100 grams (D) 1000 gramsAns: (C) 100 grams
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5.
What will the molar concentration of potassium hydroxide solution if 2.0 g of it are dissolved in 100 cm³ of solution?(A) 0.36 mol/dm³ (B) 0.53 mol/dm³ (C) 0.70 mol/dm³ (D) 0.45 mol/dm³Ans: (A) 0.36 mol/dm³ — moles = 2 ÷ 56 = 0.0357 mol; volume = 0.1 dm³; M = 0.0357 ÷ 0.1 ≈ 0.36 mol/dm³
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6.
What is the percentage of Hydrogen in Ca(OH)2?(A) 53.14% (B) 2.70% (C) 57.15% (D) 9.90%Ans: (B) 2.70% — (2 ÷ 74) × 100 = 2.70%
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7.
What will be the mass of NaCl needed to make up 15 dm³ of its solution with a concentration of 0.7 g/dm³?(A) 21 g (B) 10.5 g (C) 31.5 g (D) 5.4 gAns: (B) 10.5 g — 0.7 g/dm³ × 15 dm³ = 10.5 g
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8.
The formula which shows the simplest whole number ratio of atoms present in a compound is called:(A) Molecular formula (B) Structural formula (C) Empirical formula (D) Displayed formulaAns: (C) Empirical formula
- i.
What is the molar volume of a gas at RTP?
The volume occupied by one mole of any gas at RTP (25°C and 1 atm) is called its molar volume. Its value is 24 dm³.
- ii.
What do you know about one mole of each gas at RTP?
One mole of every gas occupies the same volume, i.e. 24 dm³ at RTP, and contains 6.022 × 1023 molecules.
- iii.
How does the molar volume relate to the Avogadro’s law?
Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. So one mole (6.022 × 1023 molecules) of any gas occupies the same volume, 24 dm³ at RTP.
- iv.
Mention the standard units for measuring solute and solution in a mixture.
Solute is measured in grams (g) or moles (mol); solution is measured in dm³ (1 dm³ = 1000 cm³). Concentration is therefore expressed as g/dm³ or mol/dm³.
- v.
Why two different compounds may show the same empirical formula?
The empirical formula gives only the simplest ratio of atoms, not the actual number. Different compounds can have the same ratio, e.g. CH2O is the empirical formula of both formaldehyde (CH2O) and glucose (C6H12O6).
- vi.
Write the formula to calculate the percentage of an element in a compound.
% of element = (Mass of element in one mole of compound ÷ Molar mass of compound) × 100
Molar volume: The volume occupied by one mole of any gas at a given temperature and pressure is called its molar volume. At RTP (25°C, 1 atm) it is 24 dm³/mol, and at STP (0°C, 1 atm) it is 22.4 dm³/mol.
Usefulness:
1. It allows us to convert the volume of a gas into moles: moles = volume ÷ 24 dm³ (at RTP).
2. It allows us to convert moles into volume: volume = moles × 24 dm³ (at RTP).
3. Volumes of gases in a reaction can be compared directly without weighing them, because equal volumes contain equal numbers of molecules.
4. Mass, moles, volume and number of particles of a gas can be interconverted easily in stoichiometric calculations.
