Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 1

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Weekly Quiz Competition

Question 1
1 / -0

We have 750 cm³ of 0.6 M Mohr's salt solution (Molar mass = 294 g), which we want to oxidise. How much potassium dichromate is required?

A
0.49 g

B
0.45 g

C
22.05 g

D
2.2 g

Solution

Oxidation number of Mohr's salt is 1.

Oxidation number of dichromate ion = 6.

Moles of Fe²⁺ = 750 * 0.6/1000 = 0.450 moles.

6 moles of Fe²⁺ = 1 mole of Cr₂O₇^{2- .}

Therefore, 0.45 mole Fe²⁺ = 0.450/6 = 0.075 mol Cr₂O₇^2-

= 0.075 * 294 g = 22.05 g
Question 2
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HCl is not used in any titration when oxalic acid is used as an reducing agent (e.g. titrating potassium solution) because ___________.

A
it gets oxidised by oxalic acid to chlorine

B
it furnishes H⁺ions in addition to those from oxalic acid

C
it reduces permanganate to Mn²⁺

D
it oxidises oxalic acid to carbon dioxide and water

Solution

Right option. Titration of oxalic acid by potassium permanganate in the presence of HCl gives unsatisfactory result because HCl is a better reducing agent than oxalic acid and HCl reduces preferably MnO₄^- to Mn²⁺.
Question 3
1 / -0

Which of the following is not an element?

A
Diamond

B
Graphite

C
Ozone

D
Helium

Solution

Ozone is a compound consisting of 3 oxygen atoms.
Question 4
1 / -0

CaO (1.88 g) is obtained from Ca (1.35 g). Use the given data to find Z (atomic weight) of Ca.

A
41 g

B
50.25 g

C
60.50 g

D
70.56 g

Solution

Moles of O = Moles of Ca (1.88-1.35)/16 = Moles of O = Moles of Ca = 1.35/Atomic weight of Ca. Atomic weight of Ca = 41 g.
Question 5
1 / -0

M₂O₃ is the formula of metal oxide. What would be the formula of its phosphate?

A
M₂PO₄

B
MPO₄

C
M₃PO₄

D
M(PO₄)₂

Solution

Oxidation number of M and phosphate is 3.
Question 6
1 / -0

0.0050 has ______ number of significant digits.

A
1

B
2

C
3

D
4

Solution

5 and 0 (right to 5) are significant digits.
Question 7
1 / -0

In compound AaBb

A
moles of A = Moles of B + Moles of AaBb

B
moles of B = Moles of A + Moles of AaBb

C
moles of AaBb = Moles of A + Moles of B

D
moles of AaBb = a* moles of A + b* Moles of B

Solution

Moles of A_aB_b = a* Moles of A + b* Moles of B.
Question 8
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What is the atomic mass of A, if 10 g of A₄O₆has 6.06 g of A?

A
32 amu

B
37 amu

C
42 amu

D
48 amu

Solution

If M is atomic mass of A, then 4M/(4M + 96 amu) = 6.06/10. Solving for M gives M = 36.9 amu.
Question 9
1 / -0

What is the average weight of a gaseous mixture of H₂and He, if mixture is equimolar in both components?

A
2.0 g

B
2.5 g

C
3.0 g

D
3.5 g

Solution

mole fraction of each component = 0.5

Average weight of the mixture = 0.5 × molar mass of H2 + 0.5 × molar mass of He

Average weight of the mixture = 0.5 × 2 + 0.5 × 4 = 3 g
Question 10
1 / -0

Number of atoms in 560 g Fe (Atomic mass 56 g/mol) is _____.

A
twice that of 70 g N

B
half that of 20 g H

C
twice that of 20 g H

D
both 1 and 2

Solution

560 g of Fe = 10 moles of Fe.
Question 11
1 / -0

Find a in SOa if density of gas is 16/5.6 g/l.

A
1

B
2

C
3

D
4

Solution

Weight of 22.4L of gas = 16*22.4/5.6 g. So for SO x= 64 g/mol 32 + 16a = 64 16a = 32 or a = 2
Question 12
1 / -0

The term representing the ratio of atoms in a compound is called ________.

A
the Empirical Formula

B
the Molecular Formula

C
the Structural Formula

D
the Rational Formula

Solution

The empirical formula of a chemical compound is the simplest whole number ratio of atoms of each element present in a compound.

However, the molecular formula gives the exact number of atoms of different elements in the compound.

e.g. The molecular formula of glucose is C₆H₁₂O₆ and the empirical formula is CH₂O.

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Some Basic Concepts of Chemistry Test - 1

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Question 1

0.49 g

0.45 g

22.05 g

2.2 g

Solution

Question 2

it gets oxidised by oxalic acid to chlorine

it furnishes H+ ions in addition to those from oxalic acid

it reduces permanganate to Mn2+

it oxidises oxalic acid to carbon dioxide and water

Solution

Question 3

Diamond

Graphite

Ozone

Helium

Solution

Question 4

41 g

50.25 g

60.50 g

70.56 g

Solution

Question 5

M2PO4

MPO4

M3PO4

M(PO4)2

Solution

Question 6

1

2

3

4

Solution

Question 7

moles of A = Moles of B + Moles of AaBb

moles of B = Moles of A + Moles of AaBb

moles of AaBb = Moles of A + Moles of B

moles of AaBb = a* moles of A + b* Moles of B

Solution

Question 8

32 amu

37 amu

42 amu

48 amu

Solution

Question 9

2.0 g

2.5 g

3.0 g

3.5 g

Solution

Question 10

twice that of 70 g N

half that of 20 g H

twice that of 20 g H

both 1 and 2

Solution

Question 11

1

2

3

4

Solution

Question 12

the Empirical Formula

the Molecular Formula

the Structural Formula

the Rational Formula

Solution

it furnishes H⁺ions in addition to those from oxalic acid

it reduces permanganate to Mn²⁺

M₂PO₄

MPO₄

M₃PO₄

M(PO₄)₂