Atoms and Molecules Test

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Atoms and Molecules Test - 37

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Question 1
1 / -0

An atom is:

A
The smallest particle of matter known

B
The smallest particle of a gas

C
The smallest indivisible particle of an element that can take part in a chemical change

D
Radioactive emission

Solution

An atom is the smallest indivisible particle of an element that can take part in a chemical change.
Greek word atom means indivisible. It is the ultimate particle of matter.
So, the correct option is $$C$$
Question 2
1 / -0

10.0 g of CaCO$$_3$$ on heating gave 4.4 g of CO$$_2$$ and 5.6 g of CaO. The observation is in agreement with the:

A
law of constant composition

B
law of multiple proportions

C
law of reciprocal proportion

D
law of conservation of mass

Solution

$$CaCO_3 \xrightarrow {\triangle} CaO + CO_2$$
$$10.0 \ g$$ $$5.6 \ g$$ $$4.4 \ g$$

Since, mass of reactant = Mass of product

$$10.0 \ g = 4.4 \ g+5.6 \ g$$

$$10.0 \ g= 10.0 \ g$$

So, the law of conservation of mass is constant.
Question 3
1 / -0

Atomic theory was given by:

A
John Dalton

B
Neils Bohr

C
E. Rutherford

D
Haber Bosch

Solution

$$(A)$$ John Dalton

$$Reason$$ : Atomic theory was given by John Dalton, represented as,
$$1$$ Everything is composed of atoms, which are the building blocks of matter and cannot be destroyed.
$$2$$ All atoms of an element are identical.
$$3$$ The atoms of different elements vary in size and mass.
$$4$$ Compounds are produced through different whole number combinations of atoms.
$$5$$ A chemical reaction results in rearrangement of atoms in the reactant and product compounds.
Question 4
1 / -0

Which of the following contains the largest number of molecules?

A
0.2 mole of H$$_2$$

B
8.0 mole of H$$_2$$

C
17 g of H$$_2$$O

D
6.0 g of CO$$_2$$

Solution

1 mole of any substance contains $$6.02 \times 10^{23 }$$ (Avogadro's number ) of molecules.

A. 0.2 mole of H$$_2$$ contains $$0.2 \times 6.02 \times 10^{23 }=1.2 \times 10^{23 }$$ molecules.

B. 8.0 mole of H$$_2$$ contains $$8.0 \times 6.02 \times 10^{23 }=4.8 \times 10^{24 }$$ molecules.

C. 17 g of H$$_2$$O $$ \displaystyle =\dfrac { \text {17 g } }{\text { 18 g/mol}} = \text { 0.94 moles}$$

17 g of H$$_2$$O contains $$0.94 \times 6.02 \times 10^{23 }=5.7 \times 10^{23 }$$ molecules.

D. 6.0 g of CO$$_2$$ $$ \displaystyle =\dfrac { \text {6.0 g } }{\text { 44 g/mol}} = \text { 0.136 moles}$$

6.0 g of CO$$_2$$ contains $$0.136 \times 6.02 \times 10^{23 }=8.2 \times 10^{22 }$$ molecules.

Hence, 8.0 mole of H$$_2$$ contains the largest number of molecules
Question 5
1 / -0

Which of the following weighs the most?

A
$$10^{23}$$ molecules of $$H_2$$

B
1 mole of $$H_2$$O

C
1 mole of $$N_2$$

D
$$10^{22}$$ molecules of $$O_2$$

Solution

Molecular weight of $$H_2=2$$ g/mol. 1 mole of $$H_2$$ weighs 2 g and contains $$6.02 \times 10^{23 }$$ (Avogadro's number ) molecules.

Weight of $$10^{23}$$ molecules of H$$_2$$ $$ \displaystyle = \dfrac {1 \times 10^{23 }}{6.02 \times 10^{23 }} \times 2=0.33$$ g

Weight of 1 mole of H$$_2$$O $$=$$ molecular weight $$ \displaystyle =2(1)+16 =18$$ g.

Weight of 1 mole of N$$_2$$ $$=$$ molecular weight $$ \displaystyle =2(14)=28$$ g.

Molecular weight of $$O_2=32$$ g/mol. 1 mole of $$O_2$$ weighs 32 g and contains $$6.02 \times 10^{23 }$$ (Avogadro's number ) molecules.

Weight of $$10^{22}$$ atoms of oxygen $$ \displaystyle = \dfrac {1 \times 10^{22 }}{6.02 \times 10^{23 }} \times 32=0.53$$ g

Hence, 1 mole of N$$_2$$ weighs the most.
Question 6
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'All matter is composed of very small particles called Parmanu' was first suggested by:

A
John Dalton

B
J.J. Thomson

C
Kanad

D
William J. Crooke

Solution

Indian philosopher who suggested that all matter is composed of very small particles was Maharshi Kanad. The particles were called Parmanu (atom).
Question 7
1 / -0

Which of the following compound has formula mass of 322 u?
$$(H= 1, C=12, 0=16, Na=23, S=32, Cu=63.5)$$

A
$$Na_2CO_3.10H_2O$$

B
$$CuSO_4.5H_2O$$

C
$$Na_2CO_3.9 H_2O$$

D
$$Na_2SO_4.10 H_2O$$

Solution

$${ Na }_{ 2 }{ CO }_{ 3 }.10{ H }_{ 2 }O=286u\\ { CuSO }_{ 4 }.{ 5H }_{ 2 }O=249.5u\\ { Na }_{ 2 }{ CO }_{ 3 }.{ 9H }_{ 2 }O=268u\\ { Na }_{ 2 }{ SO }_{ 4 }.10{ H }_{ 2 }O=322u$$
Question 8
1 / -0

2.40 g of the element $$Z$$ combines exactly with 1.6 g of oxygen to form a compound with the formula $$ZO_2$$. What is the relative atomic mass of $$Z$$?

A
24.0

B
32.0

C
48.0

D
64.0
Question 9
1 / -0

The mass of one atom of carbon-12 is:

A
$$1\ g$$

B
$$12\ g$$

C
$$12\times 6.02\times 10^{23}\ g$$

D
$$\dfrac {12}{6.02\times 10^{23}}\ g$$

Solution

One mole of any element will be equal to its molar mass or atomic weight in grams.

Atomic mass of carbon = 12 g

So, one mole of carbon atom will weigh 12 g

1 mole of atoms = $$6.02 \times 10^{23}$$ atoms

So, mass of Carbon−12 = 12 g = $$6.02 \times 10^{23}$$ atoms

Mass of 1 atom = $$\dfrac {12}{6.02 \times 10^{23}}$$ g
Question 10
1 / -0

6.4 g of oxygen will contain ______ number of oxygen molecules.

A
$$6.023 \times 10^{23}$$

B
$$6.023 \times 10^{22}$$

C
$$1.2046 \times 10^{23}$$

D
$$1.2046 \times 10^{22}$$

Solution

Molar mass of oxygen $$= 32 g/mol$$
Moles in 6.4 g oxygen $$= \dfrac{6.4}{32} = 0.2$$ moles
Number of $$O_2$$ molecules in 0.2 mole $$\displaystyle = 0.2\times 6.023\times 10^{23}$$ $$=1.2046\times 10^{23}$$ molecules

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Atoms and Molecules Test - 37

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Atoms and Molecules Test - 37

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Question 1

The smallest particle of matter known

The smallest particle of a gas

The smallest indivisible particle of an element that can take part in a chemical change

Radioactive emission

Solution

Question 2

law of constant composition

law of multiple proportions

law of reciprocal proportion

law of conservation of mass

Solution

Question 3

John Dalton

Neils Bohr

E. Rutherford

Haber Bosch

Solution

Question 4

0.2 mole of H$$_2$$

8.0 mole of H$$_2$$

17 g of H$$_2$$O

6.0 g of CO$$_2$$

Solution

Question 5

$$10^{23}$$ molecules of $$H_2$$

1 mole of $$H_2$$O

1 mole of $$N_2$$

$$10^{22}$$ molecules of $$O_2$$

Solution

Question 6

John Dalton

J.J. Thomson

Kanad

William J. Crooke

Solution

Question 7

$$Na_2CO_3.10H_2O$$

$$CuSO_4.5H_2O$$

$$Na_2CO_3.9 H_2O$$

$$Na_2SO_4.10 H_2O$$

Solution

Question 8

24.0

32.0

48.0

64.0

Question 9

$$1\ g$$

$$12\ g$$

$$12\times 6.02\times 10^{23}\ g$$

$$\dfrac {12}{6.02\times 10^{23}}\ g$$

Solution

Question 10

$$6.023 \times 10^{23}$$

$$6.023 \times 10^{22}$$

$$1.2046 \times 10^{23}$$

$$1.2046 \times 10^{22}$$

Solution

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