Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 2

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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

E
None of these

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
1 / -0

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

E
none of these

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

E
Hydrogen

Solution

Ozone is a compound consisting of 3 oxygen atoms. Correct answer.
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

E
45.65 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

E
None of these

Solution

It is correct. 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

E
5

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 AaB_b= Moles of A + Moles of B

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

E
none of these

Solution

It is correct. Moles of A_aB_b = a* Moles of A + b* Moles of B.
Question 8
1 / -0

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

E
None of these

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

Calculate the number of molecules in 1/20 g of water.

A
1.67 10²³

B
1.67 10²²

C
1.67 10²¹

D
5.02 10²¹

E
5.02 10²³

Solution

Molar mass of water = 18g
The number of molecules of water = number of moles 6.022 10²³
The number of molecules of water =
The number of molecules of water = 1.67 10²¹moles
Question 10
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

E
4.0 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 11
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

E
none of these

Solution

It is correct. 560 g of Fe = 10 moles of Fe.
Question 12
1 / -0

6.0 mg of H₂ gas can reduce 0.1596 mg of an oxide of a trivalent metal.
The molar mass of the metal is:

A
5.58

B
15.58

C
55.8

D
155.8

E
None of these

Solution
Question 13
1 / -0

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

A
1

B
2

C
3

D
4

E
None of these

Solution

Weight of 22.4 L of gas =
Weight of SO_a = 32 + 16a
64 = 32 + 16a
64 - 32 = 16a
32 = 16a
a = 2
So, the value of a in SO_a is 2
So, it is SO₂
Question 14
1 / -0

In a compound C/H = 9 and N/H = 3.5. The molar mass of the compound is 108 g mol^-1. Its molecular formula is ___________.

A
C₂H₆N₂

B
C₃H₄N

C
C₆H₈N₂

D
C₉H₁₂N₃

E
None of these

Solution

Molecular mass of the compound is 108 u.
Weight of Nitrogen atoms in one molecule of the compound = = 28 u.
Number of nitrogen atoms = 28/14 = 2.
Weight of carbon atoms in one molecule of the compound = = 72 u
Number of carbon atoms = 72/12 = 6.
Weight of Hydrogen atoms in one molecule of the compound = = 8 u
Number of Hydrogen Atom = 8/1 = 8.
So the formula is C₆H₈N_2.
Question 15
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

E
none of these

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.