Mole Concept Test

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Mole Concept Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The number of hydrogen molecules in 4.032 g is

A
6.023 × 10²³

B
2 × 6.023 × 10²³

C
4 × 6.023 × 10²³

D
4.032 × 6.023 × 10²³

Solution

Atomic mass of Hydrogen = 1.0079 u
This implies that 1 mole of Hydrogen atoms weighs 1.0079 g.
Therefore, 4.032g of hydrogen atoms will have 4.032/1.0079 mole atoms = 4 mole atoms
1 molecule of hydrogen has 2 atoms.4 moles of hydrogen atoms will be there in 2 mole molecules of hydrogen.
Hence, the number of hydrogen molecules in 4.032 g is 2 × 6.023 × 10²³.
Question 2
1 / -0

A solution is prepared by dissolving 2.6 g of H₂SO₄ in 40 g of water. Which of the following is the correct value of mole fraction of H₂SO₄?

A
0.328

B
0.126

C
85.38

D
0.012

Solution

Mass of water = 40.0 g
Molar mass of H₂O = 18 g/mol
Number of moles of water is (n₁) = Given mass/Molar mass = (40 g)/(18 g) = 2.222 mol.
Given mass of H₂SO₄ = 2.6 g
Molar mass of H₂SO₄ = 98 g
Number of moles of H₂SO₄ in the solution (n₂) = Given mass/Molar mass = (2.6 g)/(98 g) = 0.0265 mol.
Total number of moles = 2.222 + 0.0265 = 2.2485
Mole fraction of H₂SO₄(χ₂) = n₂/n_T = 0.0265/2.2485 = 0.0117 ≈ 0.012
Question 3
1 / -0

What is the ratio by mass of sulphur and oxygen in SO₂?

A
1 : 2

B
2 : 1

C
1 : 1

D
1 : 4

Solution

Sulphur dioxide (SO₂) contains one atom of sulphur and two atoms of oxygen.
Mass of one sulphur atom = 2x g
Mass of one oxygen atom = x g
Ratio of atomic masses of sulphur and oxygen in SO₂=
= = 1 : 1

Thus, the ratio of the masses of sulphur and oxygen in sulphur dioxide is 1 : 1.
Question 4
1 / -0

The number of atoms present in 112 grams of N₂ is

A
24 10²³

B
4.2 10²³

C
48.176 10²³

D
4.8 10²³

Solution

Molecular weight of N₂ = 28 g
Therefore, 112 g of N₂ is equivalent to = 4 moles of nitrogen molecules.

Therefore, 112 g of N₂ contains 4 × 6.022 × 10²³ molecules = 24.088 × 10²³ molecules or 2 × 24.088 × 10²³ atoms = 48.176 × 10²³ atoms.
Question 5
1 / -0

How many number of particles are present in 16 g of O₂?

A
6.022 ×10²³

B
3.011 ×10²³

C
12.046 ×10²³

D
3.011 ×10²²

Solution

Number of particles in 16 g of O₂ molecules = = 3.011 × 10²³
Question 6
1 / -0

Calculate the number of molecules in 4.25 g of H₂SO₄.

A
2.61 × 10²²

B
1.38 × 10²⁵

C
6.9 × 10^-24

D
3.6 × 10²³

Solution

Molecular weight of H₂SO₄ = 98 g
Therefore, 4.25 g of H₂SO₄ is equivalent to moles of H₂SO₄ molecules.
Therefore, 4.25 g of H₂SO₄contains 0.043 × 6.022 × 10²³ molecules.
= 0.261 × 10²³ molecules or 2.61 × 10²² molecules
Question 7
1 / -0

What is the mass in grams of 2.5 moles of slaked lime [Ca(OH)₂]?
(Atomic masses: Ca = 40 g, O = 16 g and H = 1 g)

A
185 g

B
74 g

C
29.6 g

D
76.5 g

Solution

Mass = Molar mass x Number of moles
Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)
= 40 + 34 = 74 g
Therefore, mass of Ca(OH)₂ = 74 x 2.5 = 185 g
Question 8
1 / -0

What is the volume (in litres) occupied by 49.8 g of HCl at STP?

A
15.99

B
20.99

C
25.99

D
30.99

Solution

22.4 l = 36 g
× 49.8 = 30.99 l
Question 9
1 / -0

Calculate the volume occupied by 2.9 g of butane at STP.

A
1.12 L

B
1.09 L

C
7.51 L

D
2.16 L

Solution

PV = nRT to this:V = nRT/P
Substitute:V = [(2.9 g / 58 g mol¯¹) (0.08206 L atm mol¯¹ K¯¹) (273.0 K)]/1.00 atm
V = 1.12 L (to three significant figures)
Question 10
1 / -0

0.1 gm of a divalent metal liberates 56 ml of hydrogen from an acid at STP. The atomic weight of the metal is

A
20

B
25

C
30

D
40

Solution

Number of moles at STP =

Equivalent Weight of gas =
Question 11
1 / -0

What is the total number of moles at the end of reaction when one mole each of CO and O₂ are made to react at STP?

A
1.5 moles

B
1 mole

C
4 moles

D
2 moles

Solution

But 1 mole of CO and 1 mole of O₂ are present. Now 1 mole of O₂ requires 2 moles of CO to react and 1 mole of CO requires 0.5 moles of O₂ to react. 2 moles of CO are not present but 0.5 moles of O₂ are present.
Therefore, 1 mole of CO will react with 0.5 moles of O₂ to form 1 mole of CO₂. 0.5 moles of O₂ will remain unreacted. In this reaction, CO is the limiting reagent because it limits the quantity of the product formed.
Finally the total moles present are 0.5 (moles of O₂) + 1 (mole of CO₂) = 1.5.
Question 12
1 / -0

Molecular weight of heavy water used as a moderator in nuclear reactors is

A
22

B
16

C
18

D
20

Solution

Molecular weight of heavy water used as a moderator in nuclear reactors is 20 g.
Question 13
1 / -0

Which of the following quantities is the molecular mass of AI₂(SO₄)₃? The atomic masses of AI, S and O are 27, 32 and 16 respectively.

A
450 g

B
342 g

C
278 g

D
273 g

Solution

The molecular mass of AI₂(SO₄)₃ is 342 g.
Question 14
1 / -0

What is the molecular mass of CaCO₃?

A
56 g

B
84 g

C
100 g

D
24 g

Solution

The molecular weight of CaCO₃is 40.0 + 12.0 + 3(16.0) = 100 g/mol.
Question 15
1 / -0

Calculate the molar mass of ammonium sulphate.

A
210 g

B
199 g

C
174 g

D
132 g

Solution

The molar mass of ammonium sulphate is 132 g/mol.
Question 16
1 / -0

How is relative molecular mass of a gas related to its vapour density?

A
Relative molecular mass of gas is equal to vapour density.

B
Relative molecular mass of gas is twice its vapour density.

C
Relative molecular mass of gas is thrice its vapour density.

D
Relative molecular mass of gas is half of its vapour density.

Solution

Vapour Density =
Or

2 × Vapour Density = Relative Molecular Mass
Relative Molecular Mass of gas is twice its Vapour Density.
Question 17
1 / -0

The molecular mass of Cl₂ is

A
70.90 u

B
17.05 u

C
89.87 u

D
98.75 u

Solution

The molecular mass of Cl₂is 70.90 u.
Question 18
1 / -0

Percentage of metal in metallic oxide is 60. Its equivalent weight is

A
(Mol. wt.)

B
(Mol. wt.)

C
(Mol. wt.)

D
(Mol. wt.)

Solution

Let the atomic weight of metal be x.

x = 24, therefore the metalic magnesium (Mg)

Equivalent weight =
So, equivalent weight of magnesium oxide = (Mol. wt.)
Question 19
1 / -0

The percentage of nitrogen in dry air is

A
88.06% by volume

B
78.09% by volume

C
68.8% by volume

D
60% by volume

Solution

Air is made up of 78 percent nitrogen, just under 21 percent oxygen and the rest is water vapour, CO2 and small concentrations of noble gases such as neon and argon.
Question 20
1 / -0

A compound of nitrogen and oxygen, on analysis, showed that it contained 28 g of nitrogen for every 32 g of oxygen. Which of the following formulae correctly represents the compound?

A
N₂O₃

B
N₂O₂

C
N₂O

D
N₂O₄

Solution

If a compound of nitrogen and oxygen, on analysis, shows that it contains 28 g of nitrogen for every 32 g of oxygen, it forms a compound with the formula N₂O₂.