31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 1 - NEET MCQ

Hence empirical formula =CH₃

Additional Information: An empirical formula represents the simplest whole-number ratio of various atoms present in a compound, whereas, the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.

We know that, 

Number of atoms = 

(a) no. of atoms

(b) no. of atoms

(c) no. of atoms

(d) no. of atoms

∴ 1 gm of Li is having maximum no. of atoms.

Millimoles of solution of chloride = 0.05 × 10 = 0.5
Millimoles of AgNO₃ solution = 10 × 0.1 = 1
So, the millimoles of AgNO₃ are double than the chloride solution

Given, the number of molecules of urea =6.02 x 10²⁰
therefore, the number of moles = 

Therefore, the correct answer is 0.01 M

• The number of moles can be calculated from the ratio of the mass to the molar mass.
• A maximum number of moles will correspond to the maximum number of molecules.

Hence, option C is correct.

1 mol of triatomic gas has 3 ×6.02×10²³ atoms

Therefore, no. of atoms in 0.1 mol =0.1×  3 ×6.02×10²³ = 1.806×10²³

Ba(OH)₂​+2HCl⟶BaCl_2​+2H_2​O

No. of milli equivalent of HCl = 20 × 0.05 = 1.0
No. of milli equivalent of Br (OH)₂   = 30 × 0.1 × 2 = 6.0
After neutralization, no. of milli equivalents in 50 ml. of solution = (6 – 1) = 5
Total volume of the solution = 20 + 30 = 50 ml
∴ No. of milli equivalent of OH^– is 5 in 50 ml, 

Therefore, the correct answer is D.

The empirical formula is to be calculated as:

Thus empirical formula is CH₃O.

Writing the equation of  combustion of propane (C₃H₈), we get

From the above equation we find that we need 5 L of oxygen at NTP to completely burn 1 L of propane at N.T.P.
If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure. The same results are obtained. i.e. 5 L is the correct answer.

No. of moles of MnO₄⁻  required to oxidize one mole of FeC_2​O_4​ completely in an acidic medium.

3MnO₄⁻ ​+ 24H⁺ + 5FeC₂​O₄​⟶ 3Mn⁺² + 12H₂​O + 10CO_{2 ​}+ 5Fe⁺³

5 moles of  FeC_2​O₄ reacts with 3 moles of MnO⁴⁻.

5 moles of FeC_2​O₄​⟶3 moles of MnO⁴⁻​

1 mole of FeC_2​O₄​⟶ 53​ moles of MnO^4−​

∴ The number of moles of MnO^4−​ required to oxidize one mole of ferrous oxalate completely in the acidic medium will be 0.6 moles.

Density of water = 1 gram cm^–3 
∴ Volume occupied by 1 gram water = 1 cm³ 
i.e. Volume occupied by  molecules of water = 1 cm³ [∵ 1g water = 1/18moles of water]

Thus volume occupied by 1 molecule of water 

The balanced chemical equation is:

From the equation, it is clear that 2 moles of MnO₄^–  are required to oxidize 5 moles of SO₃^–2  

2/5 mole of MnO₄^– is required to oxidize 1 mole of SO₃^–2 

Molarity of H₂SO₄ solution

Suppose V ml of this H₂SO₄ is used to prepare 1 lit. of 0.1M H₂SO₄
∴ V x 18.02 = 1000 x 0.1

2Al₂O₃ + 3C → Al₃CO₂

Gram equivalent  of Al₂O₃ = Gram equivalent of C

Equivalent weight of Al 

Equivalent weight of C

No. of gram equivalent of Al  = = 30 × 10³
Hence, No. of gram equivalent of C = 30 × 10³

Again, No. of gram equivalent of C
⇒  mass = 90 × 10³ g = 90 kg

It is given that only 50% of the expected product is formed.
Hence, only 10 litre of NH₃ is formed i.e. N₂ used = 5 litres ⇒ N₂ left = 30 – 5 = 25 litres
H₂ used = 15 litres ⇒ H₂ left = 30 – 15 = 15 litres

As we know, 1 mole of gas has NA​ molecules. So, the higher the number of moles of a gas, it will have higher the number of molecules.

No. of molecules in different cases:

(a) 22.4 litre at STP contains = 6.023×10²³ molecule of H₂
∴ 15 litre at STP contains molecules

(b) 22.4 litre at STP contains = 6.023×10²³ molecule of N₂
∴ 5 litre at STP contains molecules

(c) 2 gm of  H₂ = 6.023×10²³ molecules of  H₂
∴ 0.5 gm of H₂ contains molecules

(d) Similarly, 10 g of O₂ gasmolecules

Thus (a) will have maximum number of molecules

Average isotopic mass of,

= 199.96 amu ≈200 amu

2H₂ ​+ O₂​ → 2H_2​O

10 g of hydrogen (molar mass 2 g/mol) = 2 g/mol  10 g ​= 5 mol 

64 of oxygen  (molar mass 32 g/mol) = 32 g/mol  64 g ​= 2 mol 

As per the balanced chemical equation, 1 mole of oxygen will react with 2 moles of hydrogen to form 2 moles of water.

Here, oxygen is the limiting reagent and hydrogen is the excess reagent.

Hence the amount of H₂O produced depends on the amount of O₂ taken
0.5 mole of O₂ gives H₂O = 1 mol
∴ 2 mole of O₂ gives H₂O = 4 mol