Question 1Level 3 · AdvancedSingle correct
An element X forms two oxides. Oxide I contains 50% X by mass and oxide II contains 40% X by mass. The masses of oxygen that combine with a fixed mass of X in oxides I and II are in the ratio
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Answer: D
Key idea: compare oxygen per fixed mass (1 g) of X, not the percentages directly. Oxide I: 50 g X with 50 g O, so 1.0 g O per g X. Oxide II: 40 g X with 60 g O, so 1.5 g O per g X. Ratio = 1.0 : 1.5 = 2 : 3, a simple whole-number ratio as the law of multiple proportions requires (these data fit SO2 and SO3). Trap: comparing 50% and 40% of X directly gives 5 : 4; inverting gives 3 : 2. Exam pattern: NEET – law of multiple proportions from percentage compositionQuestion 2Level 3 · AdvancedSingle correct
Gay-Lussac observed that 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of water vapour at the same temperature and pressure. Dalton's atomic theory could not explain this result. Which proposal made by Avogadro removed the difficulty?
AElements can combine in more than one fixed ratio by mass to form different compounds
BMass is neither created nor destroyed in a chemical reaction
CGaseous elements such as hydrogen and oxygen exist as molecules made of more than one atom, so a molecule can split into atoms during reaction
DEqual volumes of all gases at the same temperature and pressure contain equal numbers of atoms
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Answer: C
If equal volumes contained equal numbers of atoms (Dalton's view), 1 volume of oxygen atoms could make at most 1 volume of water particles, not 2, unless an oxygen atom were split, which Dalton forbade. Avogadro distinguished atoms from molecules: oxygen gas consists of O2 molecules, so each O2 splits into two O atoms and 2 volumes of water (2H2+O2→2H2O) result. His hypothesis speaks of equal numbers of molecules, not atoms, which is why the 'equal numbers of atoms' option is the trap. The options on multiple proportions and conservation of mass are other laws that do not address volumes. Exam pattern: NEET – NCERT laws of chemical combination, Gay-Lussac and AvogadroQuestion 3Level 3 · AdvancedSingle correct
16.8 g of sodium hydrogen carbonate is heated strongly until there is no further loss in mass: 2NaHCO3→Na2CO3+H2O+CO2. Using Na = 23, H = 1, C = 12, O = 16, the mass of the residue is
A10.6 g
B12.4 g
C21.2 g
D15.0 g
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Answer: A
NaHCO3 = 84 g/mol, so 16.8 g = 0.20 mol. 2 mol NaHCO3 gives 1 mol Na2CO3, so residue = 0.10 mol x 106 = 10.6 g. Check by conservation of mass: 0.10 mol each of H2O (1.8 g) and CO2 (4.4 g) leave; 16.8 - 6.2 = 10.6 g. Trap: using a 1 : 1 ratio gives 21.2 g (more than the starting mass, which is impossible); removing only CO2 gives 12.4 g; removing only water gives 15.0 g. Exam pattern: NEET – stoichiometry of decomposition reactions