Practice high-yield Chemical Engineering MCQs on Stoichiometry — Part 1 with worked academic solutions for FPSC, PPSC, PAEC, OGDCL.
Questions
40
Difficulty Level
Exam-Level
Target Exams
FPSC, PPSC, PAEC...
Estimated Time
48 mins
Exam-Level FPSC PPSC PAEC OGDCL
Verified for Commission Syllabus • Oct 14, 2026
40
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Question 1 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The total number of atoms in 8.5 gm of NH 3 is __________ × 10 23 .
Correct Answer: Option C — 12.04
Detailed Academic Explanation:
The molecular weight of NH 3 is 17 g/mol (Nitrogen = 14, Hydrogen = 1 × 3 = 3). For 8.5 grams of NH 3 , the number of moles is calculated as: Moles of NH 3 = 8.5 / 17 = 0.5 moles. The total number of molecules in 0.5 moles is: 0.5 × 6.02 × 10 23 molecules. Since each molecule of NH 3 contains 4 atoms (1 nitrogen and 3 hydrogen), the total number of atoms is: 0.5 × 6.02 × 10 23 × 4 = 12.04 × 10 23 atoms.
💡Key Conceptual Anchor:Stoichiometry
Question 2 of 40
★ FPSC
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★ FPSC
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The temperature of a gas in a closed container is 27° C. If the temperature of the gas is incresed to 300° C, then the pressure exerted is
Correct Answer: Option A — Doubled
Detailed Academic Explanation:
According to Gay-Lussac's Law , for a gas at constant volume, the pressure is directly proportional to the absolute temperature: P ∝ T. First, convert the given temperatures to Kelvin: Initial temperature (T₁): 27°C + 273 = 300 K Final temperature (T₂): 300°C + 273 = 573 K Using the formula P₂ = P₁ × (T₂ / T₁) : P₂ = P₁ × (573 / 300) ≈ P₁ × 1.91 This shows that the final pressure is approximately 1.91 times the initial pressure, which is nearly double. Therefore, the pressure exerted is best described as doubled .
💡Key Conceptual Anchor:Stoichiometry
Question 3 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Two solutions A 1 and A 2 have pH value of 2 and 6 respectively. It implies that the solution
Correct Answer: Option B — A1 is highly acidic
Detailed Academic Explanation:
The pH scale measures the acidity or alkalinity of a solution, ranging from 0 (most acidic) to 14 (most alkaline), with 7 being neutral. A solution with pH 2 (A₁) is strongly acidic, since it is much lower than 7. A solution with pH 6 (A₂) is slightly acidic, as it is just below neutral. Therefore, A₁ is highly acidic compared to A₂. Option A is incorrect because A₁ is not alkaline at all, and Option C is incorrect because pH 2 indicates strong acidity, not slight acidity. Hence, the correct interpretation is that A₁ is highly acidic .
💡Key Conceptual Anchor:Stoichiometry
Question 4 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Simultaneous doubling of the absolute temperature of a gas and reduction of its pressure to half, will result in __________ in the volume of the gas.
Correct Answer: Option D — Four fold increase
Detailed Academic Explanation:
According to the ideal gas law , for a fixed amount of gas: PV = nRT Since the amount of gas (n) and gas constant (R) stay constant, the relationship between volume, temperature, and pressure is: V ∝ T/P You are doubling the absolute temperature (T₂ = 2T₁), and at the same time reducing the pressure to half (P₂ = 0.5P₁). So, the new volume compared to the original volume will be: V₂/V₁ = (T₂/T₁) * (P₁/P₂) Substituting the values gives: V₂/V₁ = (2T₁/T₁) * (P₁/0.5P₁) V₂/V₁ = 2 * 2 = 4 Therefore, the volume of the gas will experience a four fold increase , not just doubling.
💡Key Conceptual Anchor:Stoichiometry
Question 5 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A sample of well water contains 140 gm/m 3 Ca 2+ ions and 345 gm/m 3 Na + ions. The hardness of the sample of water, expressed in terms of equivalent CaCO 3 in gm/m 3 is (assuming atomic masses of Ca : 40, Na : 23, C : 12, O : 16)
Correct Answer: Option A — 350
Detailed Academic Explanation:
350 is the correct answer to “A sample of well water contains 140 gm/m 3 Ca 2+ ions and 345 gm/m 3 Na + ions. The hardness of the sample of water, expressed in terms of equivalent CaCO 3 in gm/m 3 is (assuming atomic masses of Ca : 40, Na : 23, C : 12, O : 16)”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 6 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The molecules of a liquid which is in equilibrium with its vapor at its boiling point on an average have equal __________ in the two phases.
Correct Answer: Option C — Kinetic energy
Detailed Academic Explanation:
Kinetic energy is the correct answer to “The molecules of a liquid which is in equilibrium with its vapor at its boiling point on an average have equal __________ in the two phases.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 7 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Which of the following holds good for a solution obeying Raoult's law (i.e., an ideal solution) (where, ΔH = heat of mixing, and ΔV = volume change on mixing ) ?
Correct Answer: Option D — Both B and C
Detailed Academic Explanation:
Both B and C is the correct answer to “Which of the following holds good for a solution obeying Raoult's law (i.e., an ideal solution) (where, ΔH = heat of mixing, and ΔV = volume change on mixing )”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 8 of 40
★ FPSC
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★ FPSC
★ PPSC
Number of gm moles of solute dissolved in 1 kg of solvent is called its
Correct Answer: Option C — Molality
Detailed Academic Explanation:
Molality is the correct answer to “Number of gm moles of solute dissolved in 1 kg of solvent is called its”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 9 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
If pH value of a solution is 8, then its pOH value will be
Correct Answer: Option A — 6
Detailed Academic Explanation:
6 is the correct answer to “If pH value of a solution is 8, then its pOH value will be”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 10 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Solution made by dissolving equimolar amounts of different solutes in the same amount of a given solvent will have the
Correct Answer: Option A — Same elevation in boiling point
Detailed Academic Explanation:
Same elevation in boiling point is the correct answer to “Solution made by dissolving equimolar amounts of different solutes in the same amount of a given solvent will have the”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 11 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
For estimation of heat capacity of a solid compound, one can use
Correct Answer: Option C — Kopp's rule
Detailed Academic Explanation:
Kopp's rule is the correct answer to “For estimation of heat capacity of a solid compound, one can use”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 12 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Dissolving a solute in a solvent does not change its
Correct Answer: Option D — None of these
Detailed Academic Explanation:
None of these is the correct answer to “Dissolving a solute in a solvent does not change its”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 13 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Heat of transition is the heat evolved or absorbed, when a substance is converted from
Correct Answer: Option D — One allotropic form to another allotropic form
Detailed Academic Explanation:
One allotropic form to another allotropic form is the correct answer to “Heat of transition is the heat evolved or absorbed, when a substance is converted from”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 14 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
__________ fuels require the maximum percentage of 'excess air' for complete combustion.
Correct Answer: Option A — Solid
Detailed Academic Explanation:
Solid is the correct answer to “__________ fuels require the maximum percentage of 'excess air' for complete combustion.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 15 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A liquid is in equilibrium with its vapor at its boiling point. On an average, the molecules in the liquid and gaseous phases have equal
Correct Answer: Option A — Kinetic energy
Detailed Academic Explanation:
Kinetic energy is the correct answer to “A liquid is in equilibrium with its vapor at its boiling point. On an average, the molecules in the liquid and gaseous phases have equal”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 16 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The vapour pressure of water is given by, in P sat = A - (5000/T), where A is a constant, Psat is the vapour pressure in atm. and T is the temperature in K.The vapor pressure of water in atm. at 50°C is approximately
Correct Answer: Option D — 0.13
Detailed Academic Explanation:
0.13 is the correct answer to “The vapour pressure of water is given by, in P sat = A - (5000/T), where A is a constant, Psat is the vapour pressure in atm. and T is the temperature in K.The vapor pressure of water in atm. at 50°C is approximately”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 17 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
An equation for calculating vapour pressure is given by, log 10 P = A - B(t + c). This is called the
Correct Answer: Option B — Antonic equation
Detailed Academic Explanation:
Antonic equation is the correct answer to “An equation for calculating vapour pressure is given by, log 10 P = A - B(t + c). This is called the”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 18 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The combustion equations of carbon and carbon monoxide are as follows: C + O 2 = CO 2 , ΔH = -394 kJ/kg.mole CO + $$\frac{1}{2}$$O 2 = CO 2 , ΔH = -284.5 kJ/kg.mole The heat of formation of CO is __________ kJ/kg.mole.
Correct Answer: Option A — -109.5
Detailed Academic Explanation:
-109.5 is the correct answer to “The combustion equations of carbon and carbon monoxide are as follows: C + O 2 = CO 2 , ΔH = -394 kJ/kg.mole CO + $$\frac{1}{2}$$O 2 = CO 2 , ΔH = -284.5 kJ/kg.mole The heat of formation of CO is __________ kJ/kg.mole.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 19 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
In case of a ternery system involving two liquid components and a solute, the ratio of the concentration of the solute in the two phases at equilibrium is called the distribution co-efficient. The distribution co-efficient depends upon the
Correct Answer: Option C — Both A & B
Detailed Academic Explanation:
Both A & B is the correct answer to “In case of a ternery system involving two liquid components and a solute, the ratio of the concentration of the solute in the two phases at equilibrium is called the distribution co-efficient. The distribution co-efficient depends upon the”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 20 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Density of carbon dioxide is __________ kg/Nm 3 .
Correct Answer: Option B — 44/22.4
Detailed Academic Explanation:
44/22.4 is the correct answer to “Density of carbon dioxide is __________ kg/Nm 3 .”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 21 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Enthalpy change resulting, when unit mass of solid is wetted with sufficient liquid, so that further addition of liquid produces no additional thermal effect, is called the heat of
Correct Answer: Option D — Complete wetting
Detailed Academic Explanation:
Complete wetting is the correct answer to “Enthalpy change resulting, when unit mass of solid is wetted with sufficient liquid, so that further addition of liquid produces no additional thermal effect, is called the heat of”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 22 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The heat capacity of a solid compound is calculated from the atomic heat capacities of its constituent elements with the help of the
Correct Answer: Option B — Kopp's rule
Detailed Academic Explanation:
Kopp's rule is the correct answer to “The heat capacity of a solid compound is calculated from the atomic heat capacities of its constituent elements with the help of the”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 23 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Equal masses of CH 4 and H 2 are mixed in an empty container. The partial pressure of hydrogen in this container expressed as the fraction of total pressure is
Correct Answer: Option B — $$\frac{8}{9}$$
Detailed Academic Explanation:
$$\frac{8}{9}$$ is the correct answer to “Equal masses of CH 4 and H 2 are mixed in an empty container. The partial pressure of hydrogen in this container expressed as the fraction of total pressure is”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 24 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The activity co-efficient of a solution, which accounts for the departure of liquid phase from ideal solution behaviour
Correct Answer: Option C — Is a function of the liquid phase composition
Detailed Academic Explanation:
Is a function of the liquid phase composition is the correct answer to “The activity co-efficient of a solution, which accounts for the departure of liquid phase from ideal solution behaviour”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 25 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
An ideal gas can be liquified, because
Correct Answer: Option C — Forces operative between its molecules are negligible
Detailed Academic Explanation:
Forces operative between its molecules are negligible is the correct answer to “An ideal gas can be liquified, because”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 26 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The value of the gas-law constant 'R' is 1.987
Correct Answer: Option D — Both B & C
Detailed Academic Explanation:
Both B & C is the correct answer to “The value of the gas-law constant 'R' is 1.987”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 27 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Volume occupied by one gm mole of a gas at S.T.P. is
Correct Answer: Option A — 22.4 litres
Detailed Academic Explanation:
22.4 litres is the correct answer to “Volume occupied by one gm mole of a gas at S.T.P. is”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 28 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The number of H + in 1 c.c solution of pH 13 is
Correct Answer: Option C — 6.023 x 107
Detailed Academic Explanation:
6.023 x 107 is the correct answer to “The number of H + in 1 c.c solution of pH 13 is”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 29 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The density of a liquid is 1500 kg/m 3 . Its value in gin/litre will be equal to
Correct Answer: Option D — 1500
Detailed Academic Explanation:
1500 is the correct answer to “The density of a liquid is 1500 kg/m 3 . Its value in gin/litre will be equal to”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 30 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
One mole of methane undergoes complete combustion in a stoichiometric amount of air. The reaction proceeds as CH 4 + 2O 2 → CO 2 + 2H 2 O. Both the reactants and products are in gas phase. ΔH° 298 = -730 kJ/mole of methane. Mole fraction of water vapour in the product gases is about
Correct Answer: Option A — 0.19
Detailed Academic Explanation:
0.19 is the correct answer to “One mole of methane undergoes complete combustion in a stoichiometric amount of air. The reaction proceeds as CH 4 + 2O 2 → CO 2 + 2H 2 O. Both the reactants and products are in gas phase. ΔH° 298 = -730 kJ/mole of methane. Mole fraction of water vapour in the product gases is about”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 31 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The boiling points for pure water and pure toluene are 100°C and 110.6°C respectively. Toluene and water are completely immiscible in each other. A well agitated equimolar mixture of toluene and water are prepared. If, at a total pressure of one standard atm. exerted by the vapours of water and toluene, the mole fraction of water X w in the vapour phase satisfies
Correct Answer: Option C — 0.5 < Xw < 1.0
Detailed Academic Explanation:
0.5 < Xw < 1.0 is the correct answer to “The boiling points for pure water and pure toluene are 100°C and 110.6°C respectively. Toluene and water are completely immiscible in each other. A well agitated equimolar mixture of toluene and water are prepared. If, at a total pressure of one standard atm. exerted by the vapours of water and toluene, the mole fraction of water X w in the vapour phase satisfies”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 32 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Pure aniline is evaporating through a stagnant air film of 1 mm thickness at 300 K and a total pressure of 100 KPa. The vapor pressure of aniline at 300 K is 0.1 KPa. The total molar concentration under these conditions is 40.1 mole/m 3 . The diffusivity of aniline in air is 0.74 × 10 -5 m 2 /s.The numerical value of mass transfer co-efficient is 7.4 × 10 -3 . The rate of evaporation of aniline is 2.97 × 10 -4 . Its units are
Correct Answer: Option D — k.mole/m2.s
Detailed Academic Explanation:
k.mole/m2.s is the correct answer to “Pure aniline is evaporating through a stagnant air film of 1 mm thickness at 300 K and a total pressure of 100 KPa. The vapor pressure of aniline at 300 K is 0.1 KPa. The total molar concentration under these conditions is 40.1 mole/m 3 . The diffusivity of aniline in air is 0.74 × 10 -5 m 2 /s.The numerical value of mass transfer co-efficient is 7.4 × 10 -3 . The rate of evaporation of aniline is 2.97 × 10 -4 . Its units are”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 33 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
A vapor whose partial pressure is less than its equilibrium vapor pressure is called a __________ vapor.
Correct Answer: Option C — Superheated
Detailed Academic Explanation:
Superheated is the correct answer to “A vapor whose partial pressure is less than its equilibrium vapor pressure is called a __________ vapor.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 34 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
One Newton is equal to __________ dynes.
Correct Answer: Option D — 105
Detailed Academic Explanation:
105 is the correct answer to “One Newton is equal to __________ dynes.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 35 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The vapor pressure of the solvent decreased by 10 mm Hg, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in vapor pressure of the solvent is required to be 20 mm Hg.?
Correct Answer: Option D — 0.6
Detailed Academic Explanation:
0.6 is the correct answer to “The vapor pressure of the solvent decreased by 10 mm Hg, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if the decrease in vapor pressure of the solvent is required to be 20 mm Hg.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 36 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The average value of heat of neutralisation of dilute solution of strong acids and strong bases is about __________ kcal/kg.mole of water formed.
Correct Answer: Option B — 13360
Detailed Academic Explanation:
13360 is the correct answer to “The average value of heat of neutralisation of dilute solution of strong acids and strong bases is about __________ kcal/kg.mole of water formed.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 37 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The heat change for the reaction, C(s) + 2S(s) → CS 2 ($$l$$), is 104.2 kJ. It represents the heat of
Correct Answer: Option A — Formation
Detailed Academic Explanation:
Formation is the correct answer to “The heat change for the reaction, C(s) + 2S(s) → CS 2 ($$l$$), is 104.2 kJ. It represents the heat of”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 38 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
Which of the following is not used for computing latent heat of vaporisation?
Correct Answer: Option D — Hess
Detailed Academic Explanation:
Hess is the correct answer to “Which of the following is not used for computing latent heat of vaporisation”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 39 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
The equilibrium value of the mole fraction of the gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid surface. This statement pertaining to the solubility of gases in liquid is the __________ law.
Correct Answer: Option B — Henry's
Detailed Academic Explanation:
Henry's is the correct answer to “The equilibrium value of the mole fraction of the gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid surface. This statement pertaining to the solubility of gases in liquid is the __________ law.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
Question 40 of 40
★ FPSC
★ PPSC
★ FPSC
★ PPSC
2 litres of nitrogen at N.T.P. weighs __________ gms.
Correct Answer: Option B — 2.5
Detailed Academic Explanation:
2.5 is the correct answer to “2 litres of nitrogen at N.T.P. weighs __________ gms.”. This question is standard in competitive examination syllabi for Stoichiometry.
💡Key Conceptual Anchor:Stoichiometry
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Chemical Engineering Basics — Part 1
Master the next curriculum module with 40 exam-standard MCQs.
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.