31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - NEET MCQ

1 mol CCl₄ vapour = 12 + 4 × 35.5 = 154 g ≡ 22.4 L at STP

∴ Density =

As we know, 22400 cc of N₂O contain 6.02 x 10²³ molecules

Since in N₂O molecule there are 3 atoms

No. of electrons in a molecule of N₂O =  7 + 1 + 8 = 22

The reactioin may given as

Z₂ O₃ + 3H₂ —→ 2Z + 3H₂O
0.1596 g of Z₂O₃ react with H₂
= 6 mg = 0.006 g
∴ 1 g of H₂ react with

26.6g of Z₂O₃

∴ Eq. wt. of Z₂O₃ = 26.6 (from the elefinition of eq. wt.) Eq. wt. of Z + Eq. wt. of O = E + 8 = 26.6

⇒  Eq. wt. of Z = 26.6 – 8 = 18.6 Valency of metal in Z₂O₃ = 3

Eq. wt.of metal 

∴ At. wt. of Z = 18.6 × 3 = 55.8

C_p / C_v = 1.4 shows that the gas is diatomic. 22.4 litre at NTP ≡ 6.02 × 10²³ molecules 11.2 L at NTP = 3.01 × 10²³ molecules
= 3.01 × 10²³ × 2 atoms = 6.02 × 10²³ atoms

C₂H₄ + 3 O₂ —→ 2CO₂ + 2H₂O 28 kg    96 kg
∵ 28 kg of C₂H₄ undergo complete combustion  by = 96 kg of O₂
∴ 2.8 kg of C₂H₄ undergo complete combustion by = 9.6 g of O₂.

6.02 × 10²³ molecules of CO =1mole of CO 6.02 × 10²⁴ CO molecules = 10 moles CO = 10 g atoms of O = 5 g molecules of O₂

4.4 g CO₂ =0.1 mol CO₂

(mol. wt. of CO₂ = 44)                  
= 6 × 10²² molecules                  
= 2 × 6 × 10²² atoms of O

Average atomic mass

Where R.A. = relative abundance
M.No = Mass number

A ccording to Avogadro's law "equal volumes of all gases contain equal numbers of molecules under similar conditions of temperature and pressure". Thus if 1 L of one gas contains N molecules, 2 L of any gas under the same conditions will contain 2N molecules.

5 MH₂SO₄ = 10 N H₂SO₄, (∵Basicity of H₂SO₄ = 2) N₁V₁ = N₂V₂,
10 × 1 = N₂ × 10 or N₂ = 1 N

No of moles of nitride ion

 = 0.3 mol 0.3 x N_A  nitride ions.
Valence electrons = 8 × 0.3 N_A = 2.4 N_A (5 + 3 due to charge). One N^3– ion contains 8 valence electrons.

On calculation we find

As the least precise number contains 3 significant figures therefore answers should also contains 3 significant figures.

Molecular weight of C₆₀H₁₂₂ = (12 × 60) + 122 = 842.
Therefore weight of one molecule

Molecular weight of ZnSO₄ .7H ₂O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287.
∴ percentage mass of zinc (Zn)

156 gm of benzene required oxygen = 15 × 22.4 litre

∴ 1 gm of benzene required oxygen

∴ 39 gm of Benzene required oxygen

As the sum of the percentage of C, H & N is 100. Thus it does not contains O atom.

Hence empirical formula = CH₄N

We know that all non -zero digits are significant and the zeros at the beginning of a number are not significant. Therefore number 161 cm, 0.161 cm and 0.0161cm have 3, 3 and 3 significant figures respectively.

Given : Percentage of the iron = 0.334%; Molecular weight of the haemoglobin = 67200 and atomic weight of the iron = 56. We know that the number of iron atoms

According to Stoichiometry they should react  as follow

Thus for 1 mole of O₂ only 0.8 mole of NH₃ is consumed. Hence O₂ is consumed completely.

(% of O in organic compound Table for empirical formula : = 100 – (40 + 6.66 ) = 53.34 % )
Empirical formula of organic compound = CH₂O.

BaCO₃ → BaO + CO₂ 
           197 gm
197 gm of BaCO₃ released carbon dioxide = 22.4 litre at STP
∴ 1 gm of BaCO₃ released carbon dioxide

∴ 9.85 gm of BaCO₃ released carbon dioxide

= 1.12 litre

Specific volume (volume of 1 gm) of cylindrical virus particle = 6.02 × 10^–2 cc/gm
Radius of virus (r) = 7 Å = 7 × 10^–8 cm
Length of virus = 10 × 10^–8 cm
Volume of virus

= 154 × 10^–23 cc

Wt. of one virus particle

∴ Mol. wt. of virus = Wt. of N_A particle

= 15400 g/mol = 15.4 kg/mole

Suppose the mol. wt. of enzyme = x
Given 100g of enzyme wt of Se = 0.5 gm

∴  In xg of enzyme wt. of Se

Hence 

∴ x = 15680 = 1.568 × 10⁴

2g of H₂ means on e mole of H₂, hence contains 6.023 × 10²³ molecules. Others have less than one mole, so have less no. of molecules.