Pre AP Chemistry -4.1D Gravimetric Analysis- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Chemistry -4.1D Gravimetric Analysis- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Chemistry -4.1D Gravimetric Analysis- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.
What is gravimetric analysis primarily based on?
A. Volume measurement
B. Temperature measurement
C. Mass measurement
D. Pressure measurement
▶️ Answer/Explanation
Ans: C
Gravimetric analysis determines the quantity of an analyte by measuring its mass after converting it into a stable, pure compound.
In precipitation gravimetry, the analyte is converted into:
A. A gas
B. A soluble salt
C. An insoluble compound
D. A colored solution
▶️ Answer/Explanation
Ans: C
The analyte is converted into an insoluble precipitate which can be filtered, dried, and weighed.
Which of the following is a requirement for a good precipitate in gravimetric analysis?
A. Very fine particles
B. Colloidal nature
C. High solubility
D. Large, pure crystals
▶️ Answer/Explanation
Ans: D
Large, well-formed crystals are easier to filter, wash, and dry. They also contain fewer adsorbed impurities compared to very fine or colloidal particles.
Which step follows precipitation in gravimetric analysis?
A. Titration
B. Filtration
C. Distillation
D. Evaporation
▶️ Answer/Explanation
Ans: B
After the precipitate forms, it must be separated from the solution by filtration before washing, drying, and weighing.
Why is digestion of a precipitate carried out in gravimetric analysis?
A. To increase the solubility of the precipitate
B. To form larger and purer crystals
C. To cool the reaction mixture
D. To change the chemical composition
▶️ Answer/Explanation
Ans: B
Digestion (heating the precipitate in its mother liquor) allows small particles to dissolve and re-form into larger crystals. This reduces surface adsorption of impurities and improves filterability.
Which of the following would cause a positive error in gravimetric analysis?
A. Loss of precipitate during filtration
B. Incomplete precipitation
C. Presence of impurities in the precipitate
D. Using insufficient sample
▶️ Answer/Explanation
Ans: C
A positive error occurs when the measured mass is higher than the true value. Impurities trapped within or adsorbed onto the precipitate increase its mass, giving a falsely high result.
During a gravimetric analysis, the precipitate was not washed properly before drying and weighing. What is the most likely effect on the final calculated concentration of the analyte?
A. The result will be lower than the true value
B. The result will be higher than the true value
C. The result will be unaffected
D. The precipitate will completely dissolve
▶️ Answer/Explanation
Ans: B
If the precipitate is not washed properly, soluble impurities remain attached. These impurities add extra mass to the precipitate, causing an overestimation of the analyte concentration (positive error).
A 0.500 g sample containing chloride ions is treated with excess silver nitrate to form AgCl precipitate.
Ag+ + Cl− → AgCl(s)
If 1.435 g of AgCl (molar mass = 143.5 g/mol) is obtained, what is the percentage of chloride in the original sample?
A. 35.5%
B. 50.0%
C. 71.0%
D. 20.0%
▶️ Answer/Explanation
Step 1: Calculate moles of AgCl
Moles = 1.435 ÷ 143.5 = 0.010 mol
Step 2: Mole ratio
1 mol AgCl contains 1 mol Cl−
So moles of Cl− = 0.010 mol
Step 3: Mass of chloride
Cl = 35.5 g/mol
Mass = 0.010 × 35.5 = 0.355 g
Step 4: Percentage of chloride
% = (0.355 ÷ 0.500) × 100 = 71.0%
Ans: C (71.0%)
50.0 cm3 of 0.200 mol/dm3 BaCl2 is mixed with 40.0 cm3 of 0.300 mol/dm3 Na2SO4.
BaCl2 + Na2SO4 → BaSO4(s) + 2NaCl
(Molar mass BaSO4 = 233 g/mol)
What mass of BaSO4 is formed?
A. 1.17 g
B. 2.33 g
C. 0.466 g
D. 0.933 g
▶️ Answer/Explanation
Step 1: Moles of BaCl2
0.200 × 0.0500 = 0.0100 mol
Step 2: Moles of Na2SO4
0.300 × 0.0400 = 0.0120 mol
Step 3: Limiting reagent
1:1 ratio → BaCl2 is limiting (0.0100 mol)
Step 4: Mass of BaSO4
0.0100 × 233 = 2.33 g
Ans: B (2.33 g)
A 1.00 g impure sample of sodium carbonate reacts with excess CaCl2 to form CaCO3 precipitate.
Na2CO3 + CaCl2 → CaCO3(s) + 2NaCl
Mass of CaCO3 formed = 0.500 g
(Molar mass CaCO3 = 100 g/mol; Na2CO3 = 106 g/mol)
What is the percentage purity of Na2CO3?
A. 50.0%
B. 53.0%
C. 62.5%
D. 75.0%
▶️ Answer/Explanation
Step 1: Moles of CaCO3
0.500 ÷ 100 = 0.00500 mol
Step 2: Mole ratio (1:1)
Moles Na2CO3 = 0.00500 mol
Step 3: Mass of pure Na2CO3
0.00500 × 106 = 0.530 g
Step 4: Percentage purity
(0.530 ÷ 1.00) × 100 = 53.0%
Ans: B (53.0%)
Determine the gravimetric factor for phosphorus (P) when precipitated as Ag3PO4.
(Ag = 107.87, P = 30.97, O = 16.00)
A. 0.0739 g P / g Ag3PO4
B. 0.239 g P / g Ag3PO4
C. 0.418 g P / g Ag3PO4
D. 0.0309 g P / g Ag3PO4
▶️ Answer/Explanation
Molar mass Ag3PO4 = (3×107.87) + 30.97 + (4×16)
= 323.61 + 30.97 + 64 = 418.58 g/mol
Gravimetric factor = 30.97 / 418.58 = 0.0739
Ans: A
A 150 mL water sample gives 0.4640 g MgO after ignition. What is the Mg concentration in g per 100 mL?
(MgO = 40.3044 g/mol, Mg = 24.305 g/mol)
A. 0.124 g/100 mL
B. 0.1865 g/100 mL
C. 0.305 g/100 mL
D. 0.464 g/100 mL
▶️ Answer/Explanation
Moles MgO = 0.4640 / 40.3044 = 0.01151 mol
Moles Mg = 0.01151 mol
Mass Mg = 0.01151 × 24.305 = 0.280 g (in 150 mL)
Per 100 mL = (0.280/150) × 100 = 0.1865 g
Ans: B
Statement I: The precipitate must have high solubility for easy separation.
Statement II: Large-particle precipitates are preferred in gravimetric analysis.
A. Both statements are correct
B. Only Statement I is correct
C. Only Statement II is correct
D. Both statements are incorrect
▶️ Answer/Explanation
Statement I is incorrect (precipitate must have LOW solubility).
Statement II is correct.
Ans: C
A 0.6128 g mixture of NaCl and KCl gives 1.039 g of AgCl upon treatment with excess AgNO3.
(AgCl = 143.32 g/mol, NaCl = 58.44 g/mol, KCl = 74.55 g/mol)
What is the percentage of K in the mixture?
A. 25.1%
B. 50.0%
C. 74.9%
D. 60.3%
▶️ Answer/Explanation
Moles AgCl = 1.039 / 143.32 = 0.00725 mol
Total moles NaCl + KCl = 0.00725 mol
Solving simultaneous equations gives moles KCl = 0.01174 mol (as derived).
Mass K = 0.01174 × 39.098 = 0.4590 g
%K = (0.4590 / 0.6128) × 100 = 74.9%
Ans: C
An unknown 1.453 g sample containing NaCl is treated with excess AgNO3 to form AgCl.
Mass of dry filter paper = 0.862 g
Mass of filter paper + precipitate (constant mass) = 3.082 g
(Molar mass AgCl = 143.32 g/mol, NaCl = 58.44 g/mol)
What is the percentage of NaCl in the unknown sample?
A. 45.8%
B. 55.0%
C. 62.4%
D. 75.2%
▶️ Answer/Explanation
Step 1: Mass of AgCl
3.082 − 0.862 = 2.22 g
Step 2: Moles of AgCl
2.22 ÷ 143.32 = 0.0155 mol
Step 3: Moles NaCl (1:1 ratio)
0.0155 mol
Step 4: Mass NaCl
0.0155 × 58.44 = 0.906 g
Step 5: Percentage
(0.906 ÷ 1.453) × 100 = 62.4%
Ans: C
In the experiment above, the precipitate was dried several times until a constant mass was obtained.
Why is obtaining a constant mass essential in gravimetric analysis?
A. To ensure all NaCl has reacted
B. To confirm complete removal of moisture
C. To increase precipitate formation
D. To reduce experimental time
▶️ Answer/Explanation
Ans: B
Constant mass ensures that all moisture and volatile impurities have been removed. This guarantees that the measured mass corresponds only to the pure, dry precipitate.
A 50.0 mL aqueous sample containing Pb2+ ions is treated with excess NaCl to form PbCl2 precipitate.
Mass of dry PbCl2 obtained = 2.350 g
(Molar mass PbCl2 = 278.1 g/mol)
What is the molarity of Pb2+ in the original solution?
A. 0.0845 M
B. 0.169 M
C. 0.338 M
D. 0.422 M
▶️ Answer/Explanation
Step 1: Moles of PbCl2
2.350 ÷ 278.1 = 0.00845 mol
Step 2: Mole ratio
1 mol PbCl2 : 1 mol Pb2+
Moles Pb2+ = 0.00845 mol
Step 3: Molarity
Volume = 50.0 mL = 0.0500 L
M = 0.00845 ÷ 0.0500 = 0.169 M
Ans: B
If the PbCl2 precipitate was not completely dried before weighing, how would this affect the calculated molarity of Pb2+?
A. The molarity would be underestimated
B. The molarity would be overestimated
C. The molarity would remain unchanged
D. The Pb2+ ions would decompose
▶️ Answer/Explanation
If the precipitate contains moisture, its mass will be artificially high.
This increases calculated moles of PbCl2, which increases calculated moles of Pb2+.
Therefore, the molarity would be overestimated.
Ans: B