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Some Basic Concepts Of Chemistry Test 14

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Some Basic Concepts Of Chemistry Test 14

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Weekly Quiz Competition

Question 1
1 / -0

What is the mass of carbon dioxide which contains the same number of molecules as are contained in 40 g of oxygen?

A
40g

B
55g

C
32g

D
44g

Solution

Molar mass of O₂ = 32 g mol^-1
32 g of O₂= 6.023 x 10²³molecules
40 g of O₂= = 7.529 x 10²³molecules
Mass of 6.023 x 10²³ molecules of CO₂ = 44 g
Mass of 7.529 x 10²³ molecules of CO₂
Question 2
1 / -0

Match the column I with column II and mark the appropriate choice.

A
(A) → (ii), (B) → (i), (C) → (iv), (D) → (iii)

B
(A) → (i), (B) → (ii), (C) → (iii), (D) → (iv)

C
(A) → (iv), (B) → (iii), (C) → (i), (D) → (ii)

D
(A) → (iv), (B) → (iii), (C) → (ii), (D) → (i)

Solution

(A): Zn + 2HCl → ZnCl₂ + H₂
1 mole of Zn produces 2 g of H₂
0.5 mole of Zn will produce 1 g of H₂
(B): C₇₀H₂₂
Molar mass - 862
Mass of atoms = 862/6.023 x 10²³ = 1.43 x 10^-21 g
(C): 70 g of CI₂ = 6.023 x 10²³ molecules
35.5 g of CI₂ = 3.01 x 10²³ molecules
(D): Molar mass of SO₂ = 64 = 1 mole
64 g of SO₂ = 6.023 x 10²³ molecules
Question 3
1 / -0

The number of oxygen atoms present in 1 mole of oxalic acid dihydrate is

A
6 x 10²³

B
6.022 x 10³⁴

C
7.22 x 10²³

D
36.13 x 10²³

Solution

One molecule of oxalic acid dihydrate (COOH)₂2H₂O contains 6 oxygen atoms.
Number of oxygen atoms in 1 mole = 6 x 6.023 x 10²³ = 36.13 x 10²³ atoms
Question 4
1 / -0

The density of a gas is 1.78 g L^-1 at STP. The weight of one mole of gas is

A
39.9 g

B
22.4 g

C
3.56 g

D
29 g

Solution

1 mole occupies a volume of 22.4 L at STP
Mass of 1 mole of a gas = Density x Volume
= 1.78 x 22.4 = 39.9 g
Question 5
1 / -0

Which of the following gases will have least volume if 10 g of each gas is taken at same temperature and pressure?

A
CO₂

B
N₂

C
CH₄

D
HCl

Solution

Number of moles ∝ 1/Molecular mass Molecular mass of CO₂ = 44, N₂ = 28, CH₄= 16, HCl = 36.5
CO₂ will have least volume, as no. of moles is directly proportional to volume at constant P and T.
Question 6
1 / -0

How many number of molecules and atoms respectively are present in 2.8 litres of a diatomic gas at STP?

A
6.023 x 10²³, 7.5 x 10²³

B
6.023 x 10²³, 15 x 10²²

C
7.5 x 10²², 15 x 10²²

D
15 x 10²², 7.5 x 10²³

Solution

Number of molecules of gas at STP
molecules
Number of atoms in diatomic molecule
= 2 x 7.5 x 10²² = 15 x 10²² atoms
Question 7
1 / -0

Total number of atoms present in 34g of NH₃ is

A
4 x 10²³

B
48 x 10²¹

C
2 x 10²³

D
48 x 10²³

Solution

No. of moles of 34 g of NH₃ = 34/17 = 2
No. of molecules = 2 x 6.023 x 10²³
No. of atoms in one molecule of NH₃ = 4
No. of atoms in 2 molecules of NH₃
= 4 x 2 x 6.023 x 10²³ = 48.18 x 10²³
Question 8
1 / -0

What will be the mass of 100 atoms of hydrogen?

A
100 g

B
1.66 x 10^-22g

C
6.023 x 10²³ g

D
100 x 6.023 x 10²³ g

Solution

Number of gram atoms of H

Mass of hydrogen atoms = 1 x 1.66 x 10^-22
= 1.66 x 10^-22 g
Question 9
1 / -0

How many atoms in total are present in 1 kg of sugar?

A
7.92 x 10²⁵ atoms

B
6 x 10²³ atoms

C
6.022 x 10²⁵atoms

D
1000 atoms

Solution

One molecule of sugar (C₁₂H₂₂O₁₁) = 45 atoms
Number of moles of sugar = 1000/342 = 2.92
Number of molecules = 2.92 x 6.023 x 10²³
= 17.60 x 10²³
molecules Number of atoms = 45 x 17.60 x 10²³
= 7.92 x 10²⁵ atoms
Question 10
1 / -0

How many moles of oxygen gas can be produced during electrolytic decomposition of 180 g of water?

A
2.5 moles

B
5 moles

C
10 moles

D
7 moles

Solution

2H₂O → 2H₂ + O₂
2 x 18 = 36g
36 g of water produces 1 mole of O₂gas.
180 g of water will produce 180/36 = 5 moles of O₂ gas.
Question 11
1 / -0

1.4 moles of phosphorus trichloride are present in a sample. How many atoms are there in the sample?

A
5.6

B
34

C
2.4 x 10²³

D
3.372 x 10²⁴

Solution

No. of atoms in 1 mole of PCI₃ = 4
No. of atoms in 1.4 moles of PCI₃
= 4 x 1.4 x 6.022 x 10²³= 3.372 x 10²⁴
Question 12
1 / -0

What will be the standard molar volume of He, if its density is 0.1784 g/L at STP?

A
11.2 L

B
22.4 L

C
5.6 L

D
2.8 L

Solution

Standard molar volume is the volume occupied by 1 mole of a gas at STP.
0.1784 g of He occupies volume = 1 L
4 g (1 mole) of He occupies 4/0.1784 = 22.4 L
Question 13
1 / -0

In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

A
1.613 x 10²³

B
6.023 x 10²³

C
1.61 x 10²⁷

D
1.61 x 10¹⁹

Solution

Volume of gas in 1 L = 0.06/100 = 6 x 10^-4L
Number of molecules of CO₂ = n x N_A
6 x 10^-4/22.4 x 6.023 x 10²³ = 1.61 x 10¹⁹
Question 14
1 / -0

What will be the weight of CO having the same number of oxygen atoms as present in 22 g of CO₂?

A
28 g

B
22 g

C
44 g

D
72 g

Solution

No. of O atoms in CO₂ = 2
Molar mass of CO₂ = 44 g
44 g = 1 mol ⇒ 22 g = 0.5 mol
1 mole of CO₂ contains = 2 x 6.023 x 10²³ O atoms
0.5 mole of CO₂ = 6.023 x 10²³ O atoms
1 mole of CO = 6.023 x 10²³ O atoms
Mass of 1 mole of CO = 12 + 16 = 28 g
Question 15
1 / -0

Match the mass of elements given in column I with the no. of moles given in column II and mark the appropriate choice.

A
(A) → (iv), (B) → (iii), (C) → (ii), (D) → (i)

B
(A) → (i), (B) → (iii), (C) → (ii), (D) → (iv)

C
(A) → (iii), (B) → (ii), (C) → (i), (D) → (iv)

D
(A) → (ii), (B) → (i), (C) → (iv), (D) → (iii)

Solution

(A) 28 gof He = 28/4 = 7mol
(B) 46 g of Na = 46/23 = 2 mol
(C) 60 g of Ca = 60/40 = 1.5 mol
(D) 27 g of Al = 27/27 = 1 mol
Question 16
1 / -0

How many number of aluminium ions are present in 0.051 g of aluminium oxide?

A
6.023 x 10²⁰ ions

B
3 ions

C
6.023 x 10²³ ions

D
9 ions

Solution

Mass of AI₂O₃ = 2 x 27 + 3 x 16 = 102
0.051 g of AI₂O₃ = 0.051/102 = 0.0005 mol
1 mol of AI₂O₃ contains 2 x 6.023 x 10²³ Al³⁺ ions
0.0005 mol of AI₂O₃ contains 2 x 0.0005 x 6.023 x 10²³
Al³⁺ ions = 6.023 x 10²⁰Al³⁺ ions
Question 17
1 / -0

Which of the following correctly represents 180 g of water?
(i) 5 moles of water
(ii) 10 moles of water
(iii) 6.023 x 10²³ molecules ofwater
(iv) 6.023 x 10²⁴ molecules of water

A
(i) and (ii)

B
(i) and (iv)

C
(ii) and (iv)

D
(ii) and (iii)

Solution

18g of H₂O = 1 mol
180 g of H₂O = 10 mol
18 g of H₂O = 6.023 x 10²³ molecules of H₂O
180 g of H₂O = = 6.023 x 10²⁴ molecules of H₂O
Question 18
1 / -0

How many grams of CaO are required to react with 852 g of P₄O₁₀?

A
852 g

B
1008 g

C
85 g

D
7095 g

Solution

6CaO + P₄O₁₀ → 2Ca₃(PO₄)₂
1 mole of P₄O₁₀ = molar mass, of P₄O₁₀ = 284g
852 g of P₄O₁₀ = 852/284 = 3mol
1 mole of P₄O₁₀ reacts with.6 moles of CaO
3 moles of P₄O₁₀ reacts with 18 moles of CaO Mass of 18 moles of CaO = 18 x 56 = 1008 g
Question 19
1 / -0

How many oxygen atoms will be present in 88 g of CO₂?

A
24.08 x 10²³

B
6.023 x 10²³

C
44 x 10²³

D
22 x 10²⁴

Solution

1 mole of CO₂ = 44 g
88 g of CO₂ = 2 moles
No. of oxygen atoms in 2 moles = 2 x 2 x N_A
= 2 x 2 x 6.023 x 10²³= 24.08 x 10²³
Question 20
1 / -0

A mixture having 2 g of hydrogen and 32 g of oxygen occupies how much volume at NTP?

A
44.8 L

B
22.4 L

C
11.2 L

D
67.2 L

Solution

2 g of H₂ = 1 mole, 32 g of O₂ = 1 mole
Total volume of 2 moles of gases at NTP = 2 x 22.4 L
= 44.8L
Question 21
1 / -0

One atom of an element weighs 3.32 x 10^-23 g. How many number of gram atoms are there in 20 kg of the element?

A
2000

B
20

C
200

D
1000

Solution

Atomic mass of an element
= Mass of one atom x N_A
= 3.32 x 10^-23 x 6.023 x 10²³ = 19.99 = 20 g
No. of gram atoms
Question 22
1 / -0

Which of the following statements about Avogadro's hypothesis is correct?

A
Under similar condition of temperature and pressure, gases react with each other in simple ratio

B
Under similar conditions of temperature and pressure, equal volume of all gases contain same number of molecules

C
At NTP all gases contain same number of molecules

D
Gases always react with gases only at the given temperature and pressure
Question 23
1 / -0

Fill in the blanks by choosing the correct options.

A
a

B
b

C
c

D
d

Solution

1 mole = 6.023 × 10²³ atoms / molecules = X
1 mole = Molar volume =22.4 L at NTP = Y
1 mole = Gram atomic mass or gram molecular mass = Z
Hence, option B is correct.
Question 24
1 / -0

The mass of one mole of a substance in grams is called its

A
Molecular mass

B
Molar mass

C
Avogadro's mass

D
Formula mass
Question 25
1 / -0

What is the total number of electrons present in 1.6 g of methane?

A
6.023 x 10²³

B
16

C
12.04 x 10²³

D
6.023 x 10²⁴

Solution

1 mole of CH₄= 16 g
16 g of CH₄ contain 10 electrons
1.6 g of CH₄ contain 10/16 x 1.6 x 6.023 x 10²³
= 6.023 x 10²³ electrons

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