NEET Some Basic Concepts of Chemistry Previous Year Questions with Solutions

In scientific calculations, bias in rounding is avoided using the round-half-to-even rule: 1. If the digit to be dropped is greater than 5, the preceding digit is increased by 1. 2. If the digit to be dropped is less than 5, the preceding digit is left unchanged. 3. If the digit to be dropped is 5 (not followed by non-zero digits): if the preceding digit is EVEN, it remains unchanged; if the preceding digit is ODD, it is increased by 1 to make it even.

• • Ensures statistical symmetry without systematic positive or negative bias.

We require 3 figures after the decimal point: 17.014[5]. The digit to be dropped is 5. The preceding digit is 4, which is an EVEN number. Therefore, according to the rule, 4 remains unchanged.

• • First rounded value: 17.014.

We require 3 figures after the decimal point: 21.023[5]. The digit to be dropped is 5. The preceding digit is 3, which is an ODD number. Therefore, according to the rule, 3 is increased by 1 to become 4.

• • Second rounded value: 21.024.
• • Resulting pair: 17.014 and 21.024 (Option 3).

Complete combustion of an alkane (C_n H_{2n+2}) with excess oxygen yields carbon dioxide and water vapor.

• • 1 mole of n-butane produces 4 moles of CO₂.

Determine the molecular mass of n-butane and calculate the number of moles in 116 g.

• • n(butane) = 116 g / 58 g mol⁻¹ = 2.0 mol.

Using stoichiometry, 2.0 moles of butane produce 8.0 moles of CO₂.

• • M(CO₂) = 12 + 2(16) = 44 g mol⁻¹.
• • Mass of CO₂ = 8.0 mol × 44 g mol⁻¹ = 352 g.
• • Matches Option (1).

John Dalton proposed his atomic theory in 1808 based on the premise that matter consists of indivisible atoms:

• • Compounds are formed by atoms combining in fixed whole-number ratios, explaining the Law of Definite (Constant) Proportion.
• • When elements form multiple compounds, atoms combine in simple integer ratios, explaining the Law of Multiple Proportion.

Gay-Lussac showed that when gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of gaseous products:

• • Because Dalton failed to distinguish between atoms and diatomic molecules, his theory could not account for fractional atomic volumes.
• • It was only in 1811 that Avogadro resolved this discrepancy with Avogadro’s hypothesis (introducing molecules).

Dalton’s theory could not explain the Law of gaseous volume, matching Option (4).

• • Matches Option (4) with 100% historical and conceptual precision.

For any substance of mass m, molar mass M, and atomicity N_{atoms} (atoms per formula unit):

• • Two samples have an equal number of atoms if their product n × N_{atoms} is identical.

Compute the moles of atoms for the solid ionic compounds:

• • B: (248 / 62) × 3 = 4 × 3 = 12 moles of atoms (12 N_A).
• • C: (240 / 40) × 3 = 6 × 3 = 18 moles of atoms (18 N_A).

Compute the moles of atoms for the gaseous species:

• • E: (220 / 44) × 3 = 5 × 3 = 15 moles of atoms (15 N_A).
• • Comparing values: A = 12 N_A, B = 12 N_A, D = 12 N_A.
• • Therefore, A, B, and D have equal numbers of atoms, matching Option (2).

Molar mass of helium is 4.00 g/mol. Calculating moles and number of atoms:

At standard temperature and pressure (T = 273.15 K, P = 1 bar), the molar volume of an ideal gas is 22.71098 L/mol:

By definition of the mole, 1 mole contains Avogadro’s number of entities:

The unified atomic mass unit u is the mass of a single nucleon scale. Because one helium-4 atom has a mass of 4.00 u:

Sodium hydroxide and hydrochloric acid undergo an equimolar 1:1 acid-base neutralization reaction:

Computing the moles of each reactant: - Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol - Initial moles of NaOH: n(NaOH) = 1.0 g / 40 g/mol = 0.025 mol = 25.0 mmol - Initial moles of HCl: n(HCl) = Molarity × Volume(L) = 0.75 mol/L × 0.025 L = 0.01875 mol = 18.75 mmol

Because the stoichiometric ratio is 1:1 and n(HCl) < n(NaOH), hydrochloric acid (HCl) is the limiting reagent and is completely consumed. Moles of NaOH reacting = 18.75 mmol. Moles of NaOH left unreacted:

Converting the remaining moles of sodium hydroxide to mass in milligrams:

The sum of percentages of all constituent elements in compound X must equal 100%:

Dividing the mass percentage of each element by its respective atomic mass: - Element A: n_A = 32 g / 64 g/mol = 0.50 mol - Element B: n_B = 20 g / 40 g/mol = 0.50 mol - Element C: n_C = 48 g / 32 g/mol = 1.50 mol

Dividing all mole numbers by the smallest mole value (0.50):

The atomic ratio 1 : 1 : 3 yields the empirical formula ABC3:

The limestone sample has a total mass of 20 g and is 20% pure. Only pure CaCO₃ decomposes according to the stoichiometry of the reaction:

Compute the molar mass of CaCO₃ and determine moles present in 4 g:

The balanced chemical equation for the thermal decomposition of limestone at 1200 K is:

Multiply the moles of CO₂ by its molar mass (M = 12 + 2 × 16 = 44 g/mol):

For a 100 g sample of the hydrocarbon: - Mass of Carbon: $w_{\text{C}} = 78\text{ g}$ - Mass of Hydrogen: $w_{\text{H}} = 100 - 78 = 22\text{ g}$

Divide the mass of each element by its respective atomic mass ($C = 12\text{ g/mol}, H = 1\text{ g/mol}$):

Divide by the smaller number of moles ($6.5$): - Atomic ratio of C: $\frac{6.5}{6.5} = 1$ - Atomic ratio of H: $\frac{22.0}{6.5} = 3.38 \approx 3$ Comparing theoretical percentages of options: - $CH$: $\%C = 12/13 \times 100 = 92.3\%$ - $CH_2$: $\%C = 12/14 \times 100 = 85.7\%$ - $CH_3$: $\%C = 12/15 \times 100 = 80.0\%$ (closest to $78\%$) - $CH_4$: $\%C = 12/16 \times 100 = 75.0\%$

For any substance of mass $$m$$: $$\text{Number of moles of molecules/atoms } n = \frac{m}{M}$$ $$\text{Total number of atoms} = n \times \text{Atomicity} \times N_A = \left(\frac{m}{M}\right) \times \text{Atomicity} \times N_A$$

1. For $$1\text{ g of Ag (s)}$$ ($$\text{Atomic mass} = 108\text{ g mol}^{-1}$$, atomicity = 1): $$\text{Atoms} = \frac{1}{108} N_A \approx 0.00926 N_A$$ 2. For $$1\text{ g of Mg (s)}$$ ($$\text{Atomic mass} = 24\text{ g mol}^{-1}$$, atomicity = 1): $$\text{Atoms} = \frac{1}{24} N_A \approx 0.04167 N_A$$ 3. For $$1\text{ g of } \text{O}_2\text{ (g)}$$ ($$\text{Molar mass} = 32\text{ g mol}^{-1}$$, atomicity = 2): $$\text{Atoms} = \left(\frac{1}{32}\right) \times 2 \times N_A = \frac{1}{16} N_A \approx 0.06250 N_A$$ 4. For $$1\text{ g of Li (s)}$$ ($$\text{Atomic mass} = 7\text{ g mol}^{-1}$$, atomicity = 1): $$\text{Atoms} = \frac{1}{7} N_A \approx 0.14286 N_A$$

$$0.14286 N_A > 0.06250 N_A > 0.04167 N_A > 0.00926 N_A$$ Clearly, 1 g of lithium possesses more than double the number of atoms compared to 1 g of oxygen gas, and over 15 times that of silver.

Option (4) contains the maximum number of atoms.