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Some Basic Concepts of Chemistry · Question 5

NEET (UG) 2025MCQEasy📈 Topic in 50% papers+4−1
1

Among the following, choose the ones with equal number of atoms. Choose the correct answer from the options given below:

  1. B, C, and D only
  2. B, D, and E only
  3. A, B, and C only
  4. A, B, and D only

Answer

(4) A, B, and D only

Solution

Moles of atoms = moles of substance × atoms per formula unit:

A. 212106 = 2 mol × 6 = 12 mol atoms

B. 24862 = 4 mol × 3 = 12 mol atoms

C. 24040 = 6 mol × 3 = 18 mol atoms

D. 122 = 6 mol × 2 = 12 mol atoms

E. 22044 = 5 mol × 3 = 15 mol atoms

A, B and D have equal numbers of atoms (12 NA).

Answer: (4)

NEET (UG) 2025

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2

Dalton's Atomic theory could not explain which of the following?

  1. Law of multiple proportion
  2. Law of gaseous volume
  3. Law of conservation of mass
  4. Law of constant proportion
NEET (UG) 2025
3

1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

  1. 750 mg
  2. 250 mg
  3. Zero mg
  4. 200 mg
NEET (UG) 2024
4

The highest number of helium atoms is in

  1. 4 mol of helium
  2. 4u of helium
  3. 4 g of helium
  4. 2.271098 L of helium at STP
NEET (UG) 2024
5

A compound X contains 32\% of A, 20\% of B and remaining percentage of C. Then, the empirical formula of X is : (Given atomic masses of A=64 ; B=40 ; C=32 u)

  1. A2BC2
  2. ABC3
  3. AB2C2
  4. ABC4
NEET (UG) 2024
6

The amount of glucose required to prepare 250 mL of M20 aqueous solution is : (Molar mass of glucose : 180 g mol-1)

  1. 2.25 g
  2. 4.5 g
  3. 0.44 g
  4. 1.125 g
RE-NEET (UG) 2024
7

1.0 g of H2 has same number of molecules as in:

  1. 14 g of N2
  2. 18 g of H2O
  3. 16 g of CO
  4. 28 g of N2
RE-NEET (UG) 2024
8

The right option for the mass of CO2 produced by heating 20 g of 20\% pure limestone is (Atomic mass of Ca = 40) [ CaCO3 1200 K CaO + CO2]

  1. 1.76 g
  2. 2.64 g
  3. 1.32 g
  4. 1.12 g
NEET (UG) 2023
9

In one molal solution that contains 0.5 mole of a solute, there is

  1. 500 mL of solvent
  2. 500 g of solvent
  3. 100 mL of solvent
  4. 1000 g of solvent
NEET (UG) 2022
10

What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + 2H2O(l) [Calculate upto second place of decimal point]

  1. 1.25 g
  2. 1.32 g
  3. 3.65 g
  4. 9.50 g
NEET (UG) 2022
11

The density of the solution is 2.15 g mL-1, then mass of 2.5 mL solution in correct significant figures is :

  1. 53.75 g
  2. 5375 10-3 g
  3. 5.4 g
  4. 5.38 g
NEET (UG) 2022 (Phase 2)

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