NEET Chemistry: Quantitative Analysis Practice Questions
11 free multiple-choice questions on Quantitative Analysis for NEET Chemistry, each with the correct answer and a step-by-step explanation. Attempt a question, then reveal the solution to check your reasoning.
Practice thousands more Quantitative Analysis questions in the app →Q1. If the relative atomic mass of an element is 32, what does this value primarily represent in the context of elemental estimation?
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Correct answer: A — The mass of one mole of atoms of the element
Relative atomic mass (or atomic mass) indicates the average mass of one mole of atoms of that element, serving as the conversion factor between mass and the number of moles in calculations.
Q2. In a chemical reaction, if reactant A reacts with reactant B in a 1:2 molar ratio, which reactant is the limiting reagent?
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Correct answer: A — Reactant A
The limiting reagent is the reactant that gets completely consumed first. If the ratio requires A to react completely with B (1:2), A is the limiting reagent.
Q3. Which of the following is the most crucial step before estimating the mass of an unknown element in an analytical process?
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Correct answer: B — Determining the purity of the sample
Before any estimation can be accurate, the purity of the sample must be established. Impurities directly affect the final calculated elemental mass.
Q4. In elemental analysis, the stoichiometric ratio derived from the chemical formula is used to determine the relative amounts of elements. This ratio is based on the concept of:
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Correct answer: A — The Law of Multiple Proportions
The Law of Multiple Proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. This underpins stoichiometric ratios.
Q5. In a compound of Mg(OH)₂, the mass percentage of Magnesium is 24.3% and the mass percentage of Oxygen is 75.7%. What is the empirical formula of the compound?
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Correct answer: C — Mg(OH)₂
To find the empirical formula, we first find the atomic masses of Mg (24) and O (16). Mass of Mg = 24.3; Mass of O = 75.7. Relative number of moles of Mg = 24.3/24 = 1.0125. Relative number of moles of O = 75.7/16 = 4.73. The ratio 1.0125 : 4.73 is approximately 1:4, leading to the empirical formula Mg(OH)₂ (or Mg(OH) if simplifying the ratios based on the known formula).
Q6. The mass percentage of an element in a compound is calculated by dividing the mass of the element by the total mass of the compound and multiplying by 100. What is the correct formula?
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Correct answer: C — (Mass of Element / Mass of Compound) * 100
Mass percentage is defined as (Mass of the specific element / Total mass of the compound) * 100.
Q7. If 1 mole of a gas occupies a volume of 22.4 L at STP, what is the volume occupied by 3 moles of the same gas?
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Correct answer: D — 44.8 L
Since the volume is directly proportional to the number of moles (at constant temperature and pressure), 3 moles will occupy 3 times the volume: 3 * 22.4 L = 67.2 L.
Q8. Which of the following compounds has the empirical formula CH₂O?
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Correct answer: C — H₂O
For H₂O, the ratio of H to O is 2:1. Dividing by the smallest whole number gives 2:1. Thus, the empirical formula is H₂O.
Q9. If the density of a metal is 10 g/cm³ and its molar mass is 50 g/mol, what is the mass of 1 mole of this metal?
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Correct answer: C — 100 g
Mass of 1 mole is equal to the molar mass. Given molar mass is 50 g/mol, the mass of 1 mole is 50 g. (Note: Density is extra information not needed for this specific direct calculation).
Q10. The empirical formula of a compound of calcium nitrate (Ca(NO₃)₂) is Ca(NO₃)₂. If the molecular mass of the compound is 164 g/mol, what is the molecular formula?
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Correct answer: D — Ca(NO₃)₂
The molecular mass of Ca(NO₃)₂ is calculated as: Ca(40) + 2 × [N(14) + 3 × O(16)] = 40 + 2 × (14 + 48) = 40 + 2 × 62 = 164 g/mol. Since the calculated molecular mass matches the given molecular mass, the molecular formula is the same as the empirical formula, Ca(NO₃)₂.
Q11. If a compound contains 40% Carbon, 6.67% Hydrogen, and 53.33% Oxygen by mass, what is its empirical formula?
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Correct answer: C — CH2O
To find the empirical formula, assume a 100g sample. C = 40g/12 = 3.33; H = 6.67g/1 = 6.67; O = 53.33g/16.67 = 3.2. The simplest ratio is 3:1:2, leading to the empirical formula C3H2O. (In simplified terms, using the provided percentages directly leads to the ratio 3:1:2).
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