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


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25 Questions MCQ Test Chemistry Class 11 - 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 for NEET 2024 is part of Chemistry Class 11 preparation. The 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 questions and answers have been prepared according to the NEET exam syllabus.The 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 below.
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31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 1

At S.T.P. the density of CCl4 vapours in g/L will be nearest to: [1988]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 1

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

∴ Density =

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

One litre hard water contains 12.00 mg Mg2+.Mili-equivalents of washing soda required to remove its hardness is: [1988]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 2

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 3

1 c.c. N2O at NTP contains : 

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 3

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 4

A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is [1989]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 4

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 5

Ratio of Cp and Cv of a gas ‘X’ is 1.4. The number of atoms of the gas ‘X’ present in 11.2 litres of it at NTP will be [1989]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 5

Cp / Cv = 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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 6

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 6

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.

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 7

The number of gram molecules of oxygen in 6.02 × 1024 CO molecules is [1990]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 7

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 8

The number of oxygen atoms in 4.4 g of CO2 is

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 8

4.4 g CO2 =0.1 mol CO2

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 9

Boron has two stable isotopes, 10B (19%) and 11B (81%). Average atomic weight for boron in the periodic table is [1990]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 9

Average atomic mass

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 10

The molecular weight of O2 and SO2 are 32 and 64 respectively. At 15°C and 150 mm Hg pressure, one litre of O2 contains ‘N’ molecules. The number of molecules in two litres of SO2 under the same conditions of temperature and pressure will be : [1990]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 10

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.

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 11

A 5 molar solution of H2SO4 is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be : [1991]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 11

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 12

If NA is Avogadro’s number then number of valence electrons in 4.2g of nitride ions (N3–) is

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 12

No of moles of nitride ion

 = 0.3 mol 0.3 x NA  nitride ions.
Valence electrons = 8 × 0.3 NA = 2.4 NA (5 + 3 due to charge). One N3– ion contains 8 valence electrons.

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 13

In the final answer of the expression

the number of significant figures is : [1994]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 13

On calculation we find

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 14

The weight of one molecule of a compound C 60 H122 is [1995]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 14

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 15

The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 15

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 16

Liquid benzene (C6H6) burns in oxygen according to the equation 2C6H6(l ) + 15O2(g) —→ 12CO2(g)  6H2O(g) How many litres of O2 at STP are needed to complete the combustion of 39 g of liquid benzene?(Mol. wt. of O2 = 32, C6H6 = 78)  [1996]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 16

156 gm of benzene required oxygen = 15 × 22.4 litre

∴ 1 gm of benzene required oxygen

∴ 39 gm of Benzene required oxygen

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 17

An  organic compound containing C, H and N gave the following analysis : C = 40% ; H = 13.33% ; N = 46.67%Its empirical formula would be [1998]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 17

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

Hence empirical formula = CH4N

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 18

The number of significant figures for the three numbers 161 cm, 0.161 cm, 0.0161 cm are[1998]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 18

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.

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 19

Haemoglobin contains 0.334% of iron by weight.The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe is 56) present in one molecule of haemoglobin are [1998]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 19

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 20

In the reaction 4 NH3 (g) + 5O2(g) → 4NO(g) + 6H2O(l) When 1 mole of ammonia and 1 mole of O2 are made to react to completion, [1998]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 20

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.

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 21

An organic compound containing C, H and O gave on analysis C – 40% and H – 6.66%. Its empirical formula would be [1999, 94]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 21

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 22

Assuming fully decomposed, the volume of CO2 released at STP on heating 9.85 g of BaCO3 (Atomic mass, Ba = 137) will be [2000]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 22

BaCO3 → BaO + CO
           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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 23

Specific volume of cylindrical virus particle is 6.02 × 10–2 cc/gm. whose radius and length 7 Å & 10 Å respectively. If NA = 6.02 × 1023, find molecular weight of virus [2001]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 23

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 NA particle

= 15400 g/mol = 15.4 kg/mole

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 24

Percentage of Se in per oxidase anhydrase enzyme is 0.5% by weight (at. wt. = 78.4) then minimum molecular weight of peroxidase anhydrase enzyme is [2001]

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 24

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

31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 25

Which has maximum number of molecules?

Detailed Solution for 31 Year NEET Previous Year Questions: Some Basic Concepts Of Chemistry - 2 - Question 25

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.

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