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Resonance

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November 29, 2025
Sigma & Pi Bond

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November 29, 2025
VBT (Valence Bond Theory)

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November 29, 2025
VSEPR Theory Part-2

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November 29, 2025
VSEPR Theory Part-1

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November 29, 2025
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November 28, 2025
Shapes of Molecule

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November 28, 2025
Mole Concept Part-2
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Mole Concept Part-2

1 / 48

Total number of protons in 10 g of calcium carbonate is:

3.01×10²⁴

4.06×10²⁴

2.01×10²⁴

3.24×10²⁴

Explanation:

2 / 48

total number of electrons present in 1.4 g of dinitrogen gas.

4.2154×10²³ electrons.

4.2154×10²² electrons.

4.2154×10²⁴ electrons.

None

Explanation:

3 / 48

The total number of electrons in 18 ml of water (density $= 1 g m l^{-1})$ is :

6.02×10²³

6.02×10²⁵

6.02×10²⁴

6.02×18×10²³

The total number of electrons in $18$ ml of water is $6.02 \times 1024$ . Since, the density is $1$ g/mL, $18$ ml of water corresponds to $18$ g or $1$ mole as the molecular weight of water is $18$ g/mole.

$1$ mole of water contains $10$ moles of electrons, $8$ moles of electrons from $O$ and $2$ moles of electrons from $H$ .

$1$ mole of electrons corresponds to Avogadro’s number which is $6.02 \times 1023$ $\times$ $10$ moles of electrons will correspond to $6.02 \times 1024$ electrons.

4 / 48

The volume of oxygen evolved at STP, by decomposition of 0.68 g ‘20 volume’ hydrogen peroxide solution, is

2.24 mL

22.4 mL

224 mL

2240 mL

Explanation:

2 H₂O₂ → 2 H₂O + O₂

Moles of H₂O₂ in 0.68 g:
Molar mass of H₂O₂ = 34 g/mol
n(H₂O₂) = 0.68 / 34 = 0.02 mol

From the equation,
2 mol H₂O₂ → 1 mol O₂
So 0.02 mol H₂O₂ → 0.02 / 2
= 0.01 mol O₂

Volume of 0.01 mol O₂ at STP:
1 mol gas at STP = 22.4 L
= 22400 mL
Volume = 0.01 × 22.4 L
= 0.224 L
= 224 mL

So, the volume of O₂ evolved at STP = 224 mL.

5 / 48

Volume occupied by one molecule of water (density = 1 g cm^–3) is

9.0 × 10^–23 cm3

6.023 × 10^–23 cm3

3.0 × 10^–23 cm3

5.5 × 10^–23 cm3

Given

Density of water (ρ) = 1g/cm³

Molar mass of water (M) = 18 g/mol

∴ Molar volume (Vm) = M/ρ

V_m = 18 cm³/mol

∴ Volume of 1 molecule of water = V_m/N_A

Where N_A is Avogadro number

Volume of 1 molecule of water = 18/(6.022 × 10²³)

Volume of 1 molecule of water = 2.989 × 10^-23 ≈ 3 × 10^-23.

6 / 48

1 mole of ammonia is of 17 g. Then what is the mass of 0.3 moles of ammonia?

21g

5.1g

17g

1g

The mass of 0.3 moles of ammonia can be calculated using the following steps:

The molar mass of ammonia is 17 grams/mole. Therefore, the mass of 0.3 moles of ammonia is (0.3 moles * 17 grams/mole) = 5.1 grams.

7 / 48

The volume of occupied by 0.25 moles of H2 gas is at RTP is…..

24dm3

22.4dm3

6dm3

5.6dm3

One mole H2 bas occupy 24-liter volume at RTP hence 0.25 moles of H2 occupy.
0.25×24L. =6 Liter

NOTE:
Question Me RTP he STP hota then Answer 5.6 hota

8 / 48

The number of moles in 56gnof Fe …….

1

2

3

4

mass in gram/molar mass

9 / 48

correct formula for Calculating the number of moles is/are……

mass in grams/mol.mass

number of particles/NA

volume in dm3/volume at STP

all of these

all these formula are used for calculating number of moles.

10 / 48

The number of moles in 36g of water are …..

1moles

2 moles

3moles

4 moles

number of moles= mass in grams/molar mass

11 / 48

The number of moles in one liter of water are…..

1mole

555moles

55.5 moles

1000moles

Because 1litre=1kg
1kg water will have mass 1000 grams so
Moles=mass/molar mass
=1000/18 = 55.5 moles

12 / 48

How much mass of O2 is required to form 0.25 moles of CO2.

32g

16g

8g

4g

C(s)+O2(g)➡️CO2.One mole of CO2 requires 32 gram of O2 so 0.25 moles will require 8g of O2

13 / 48

The volume of 1gram molecular mass of CO2 at RTP is …….

44dm3

44g

22.4dm3

24dm3

At RTP one mole of each gas occupies 24dm3 of volume.

14 / 48

The volume occupied by 1/4 moles of SO2(g) at RTP is….

22.24dm3

5.6 dm3

6 dm3

2.50dm3

1 mole= 24
1/4 moles = x
Cross Multiplication
so
x=24/4
x=6

15 / 48

The maximum No. of moles are present in

11.2 dm3 of H2, gas at STP

44.8 dm3 of N2, gas at STP

22.4 dm3 of O2,gas at STP

67.2 dm3 of CO2, gas at STP

16 / 48

Which of the following is not true for a mole?

it is a counting unit

it is the gram atomic or gram formula mass of a substance

it contains different number of particles for different substances

It contains 6.023×10 particles

Explanation:

Statement C is incorrect because, by definition, one mole of any substance contains the same number of particles (Avogadro’s number), regardless of what the substance is. The mass of that mole, however, will be different for different substances.

17 / 48

if four moles of SO2 are oxidized to SO3 how many moles of oxygen molecules are required

2.0

1.5

1.0

0.5

18 / 48

Number of moles of 1 $m 3$ gas, at STP, are:

44.6

40.6

42.6

48.6

Explanation:

At STP , volume of 1 mole $=$ 22.7 L

1 $m 3$ has 1000 L
Then, n (no. of moles) in 1 $m 3 = 1000 / 22.7 =$ $44.642 m o le$

19 / 48

the ratio of no. of moles of reactants taking part and the no. of moles of product formed is called

mole ratio

mass ratio

any of these

none of these

The ratio of no. of moles of reactants taking part and the no. of moles of product formed is called mole ratio

20 / 48

Which of the following weighs the most?

One mole of sodium

One mole of water

One g – atom of nitrogen

none

Explanation:

The molar mass of sodium (Na) is its atomic mass expressed in grams per mole. The atomic mass of sodium is approximately 23 g/mol.
Mass of one mole of sodium=23 g

21 / 48

22414cm³ of an ideal gas is equal to

1 mole

2 mole

3 mole

none

22,414 cm³ of an ideal gas is approximately equal to 1 mole at standard temperature and pressure (STP).

22 / 48

number of moles of water present in 100 g of water are

5.6

7

10

none

23 / 48

the mass of one mole of electron is

1.008mg

0.55mg

0.1

none

24 / 48

one mole of any gas at STP occupies a volume of ___________

22.4dm3

24dm3

25dm3

none

Explanation:

At standard temperature and pressure (STP), one mole of any ideal gas occupies a volume of 22.4 dm3

.

25 / 48

the ratio of no. of moles of reactants taking part and the no. of moles of product formed is called

mole ratio

mass ratio

any of these

none of these

The ratio of no. of moles of reactants taking part and the no. of moles of product formed is called mole ratio

26 / 48

Which of the following weighs the most?

One mole of sodium

One mole of water

One g – atom of nitrogen

none

1 mole of Na=23g

27 / 48

22414cm³ of an ideal gas is equal to

1 mole

2 mole

3 mole

none

Explanation:

A fundamental concept in chemistry, derived from the ideal gas law (PV = nRT) and Avogadro’s law, is that one mole of any ideal gas at standard temperature 0C or273.15 K occupies a standard molar volume of approximately 22.4 liters or more precisely22.414L

28 / 48

number of moles of water present in 100 g of water are

5.6

7

10

none

Explanation:

n=100/18=5.6 mol

29 / 48

when we compare one mole each of O2 and H2 then we observe

same number of molecules

N2 has greater number of molecules than O2

both

none

Explanation:

This is due to the definition of a mole and Avogadro’s number: one mole of any substance contains a fixed number of constituent particles (molecules, in this case)

30 / 48

the number of oxygen in 0.5 mole of Al2(CO3)3 is

4.5×10^23

3.6 x10^24

2.7 x10^24

9.0 x10^23

EXPLANATION:

no. of atoms of oxygen= n ×NA× no. of atoms in formula

no. of atoms of oxygen = 0.5 × 6.023 × 10^23 ×9

no. of atoms of oxygen= 27.09 × 10^23

no. of atoms of oxygen=2.7 × 10^24

31 / 48

calculate the volume occupied by 2.8g of nitrogen gas at STP

22.4dm3

2.24 dm3

4.48 dm3

44.8 dm3

EXPLANATION:

no. of moles= volume in dm3/ molar volume

volume in dm3= no. of moles × molar volume

first find no. of moles = n= mass in grams/ molar mass

n=2.8/28

n=0.1 moles

now putting in formula

volume in dm3= n × molar volume

volume in dm3= 0.1 × 22.4

volume in dm3= 2.24 dm3

32 / 48

one mole is the amount of substance which contains as many elementary entities as contained in

0.12 kg

1.2 kg of C12 atom

0.012 kg of C12 atom

0.12 kg of O16

EXPLANATION:

one mole of C-12=12g

and 12 g= 0.012 kg

33 / 48

calculate the no. of moles of NaCl in 75.0g of table salt

0.643

0.779

28.0

1.28

EXPLANATION:

No. of moles= n= mass in grams/ molar mass

n= 75/58= 1.29

34 / 48

one mole of which one of the following bucky ball will have more molecules

C20

C50

C60

all same

EXPLANATION:

as one mole is a constant number so number of molecules is same for all of them

35 / 48

the atomic mass, molecular mass, formula mass or ionic mass expressed in grams is called as

relative atomic mass

molar mass

mole

avogadro’s number

the atomic mass, molecular mass, formula mass or ionic mass expressed in grams is called as mole

36 / 48

one gram atom of an element contains

the same number of atoms

greater number of atoms

lesser number of atoms

none

one gram atom of an element contains the same number of atoms, which is Avogadro’s number (approximately 6.022 × 10²³ atoms). This is true for any element, as one gram atom refers to the amount of an element that equals its atomic mass in grams, ensuring it contains the same number of atoms.

37 / 48

1 mol of oxygen contains

16g of oxygen

6.022×10^23 atoms of oxygen

both

none

1 mole of oxygen atoms contains:

16 grams of oxygen (since the atomic mass of oxygen is 16 g/mol).

6.022 × 10²³ atoms (Avogadro’s number).

38 / 48

the amount of a substance which contains avogadro’s number of particles is called as

molar mass

formula mass

atomic mass

mole

the amount of a substance which contains avogadro’s number of particles is called as mole

39 / 48

mass in grams of one mole of a substance is called as

ionic mass

formula mass

molar mass

none

mass in grams of one mole of a substance is called as molar mass

40 / 48

the term molar mass is also used to the mass of __________of any substance

mole

atoms

molecules

all of these

The term molar mass is also used to refer to the mass of one mole of any substance, including moles, atoms, and molecules. Molar mass is typically expressed in grams per mole (g/mol) and provides a convenient way to convert between the mass of a substance and the amount of substance in moles.

41 / 48

which one of the following is correct

n=mass in grams/molar mass

molar mass= mass in grams × n

mass in grams= n×molar mass

Both A and C

both statements are correct.

The equation
𝑛=mass in grams/molar mass

calculates the number of moles (𝑛) of a substance based on its mass and molar mass.

The equation
mass in grams=n×molar mass calculates the mass of a substance when you know the number of moles and the molar mass.

42 / 48

which one is correct

n=mass in grams/molar mass

n= number of particles/particles per mole

n= number of particles/ avogadro’s number

all of these

43 / 48

how many moles are there in 60g of sodium hydroxide

1.5 mol

2.5 mol

3.5 mol

4.5 mol

44 / 48

what is the mass of 0.5 moles of calcium carbonate (CaCO3)

25 g

50 g

100 g

150 g

Molar Mass of CaCo3=100 g/mol
mole=Mass/Molar Mass
0.5=mass/100
mass=50 g

45 / 48

one mole of any gas at standard temp and pressure occupies a volume of 22.4 dm³ this is called as

molar mass

molar volume

moles

none

one mole of any gas at standard temp and pressure occupies a volume of 22.4 dm³ this is called as molar volume

46 / 48

molar volume of a gas at STP is equals to

24 dm^3

22.4 dm^3

20 dm^3

none

molar volume of a gas at STP is equals to 22.4dm³

47 / 48

molar volume of a gas at RTP is equals to

20 dm^3

22.4 dm^3

24 dm^3

none

RTP refers to temperature of 25°C (298K) and pressure of 1 atmosphere. The molar volume of any gas at RTP is 24.0 dm³ ( 24.0 L).

48 / 48

which one is correct

n=volume dm^3 at STP/ molar volume at STP

n =volume dm^3 at STP ×molar volume at STP

n=volume dm^3 at STP+molar volume at STP

none

Your score is
The average score is 2%
November 28, 2025
CB

Category: Chemical Bonding Topical

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November 28, 2025
Limiting & Excess Reagent Part-2

Wrong shortcode initialized

November 28, 2025