Atoms and Molecules Test

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Atoms and Molecules Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The formula unit mass of $$Ca_3 (PO_4)_2$$ is :

A
$$360$$

B
$$240$$

C
$$118$$

D
$$310$$

Solution

Mass of $$Ca_3 (PO_4)_2$$
$$= (3 \times $$ atomic mass of Ca$$) + (2 \times$$ atomic mass of phosphorus$$) + (8 \times $$atomic mass of oxygen) $$= 3\times 40+2\times31+8\times16 = 310$$
Question 2
1 / -0

Which of the following obeys the law of constant proportions in their formation?

A
Mixtures

B
Compounds

C
Elements

D
Colloids

Solution

The Law of constant proportion is the rule where elements combine in a fixed proportion by mass.
Thus compounds are the kind that is formed from a fixed proportion of elements.
For example, Water $$\rightarrow H_2O$$ is obtained from any source of water like a river, lake, pond. Here the ratio of $$H$$ and $$O$$ will be always $$2:1$$.
Hence, option B is correct.
Question 3
1 / -0

The number of atoms in one mole of a substance is equal to ................ ?

A
Avagadro number

B
Mass number

C
Both (a) and (b) above

D
Density

Solution

Avagadro's number is the number of units in one mole of any substance, equal to $$6.023\times10^{23}$$. The units may be electrons, ions or molecules depending on nature of the substance & character of the relation.
$$\Rightarrow$$ If one mole$$=[6.023\times10^{23}]$$, then number of atoms is Avagadro's number.
Question 4
1 / -0

Rahul dropped 150 g of marble chips into 1 kg of HCl solution containing 0.1 of its weight of pure acid. The mass of marble chips that remained undissolved will approximately be:

A
6.50 g

B
13 g.

C
26 g.

D
15 g.
Question 5
1 / -0

Calculate the molecular mass of the caustic potash :

A
$$K_2OH$$ $$ = 95 g mol^{1}$$

B
$$ KOH = 56 g mol^{1}$$

C
$$K(OH)_2$$ $$ = 73 g mol^{1}$$

D
$$ KOH = 58 g mol^{1}$$

Solution

Potassium has valency $$ +1$$ and is represented as $$K^+$$, hydroxyl has valency $$-1$$ and is represented as $$OH^-$$. Caustic potash is $$KOH.$$ Molar mass - $$(39 +16+1) = 56 g mol^{-1}$$
Question 6
1 / -0

The molecular weight of calcium carbonate $$CaCO_{3}$$ is:

A
80

B
90

C
100

D
110

Solution

Molecular weight is the mass of whole molecule.
Steps:- (1) Chemical formula of calcium carbonate $$\longrightarrow CaCO_3$$
(2) Mass of Calcium$$\longrightarrow 40$$
Mass of Carbon$$\longrightarrow 12$$
Mass of Oxygen$$\longrightarrow 16$$
(3) Multiply mass of each element by number of atoms of that element.
(4) Add all the masses obtained after multiplication.
$$\longrightarrow CaCO_3$$
$$=1\times 40+1\times 12+3\times 16$$
$$=40+12+48$$
$$=100$$.
Question 7
1 / -0

The weight of $$100$$ molecules of each of the following substances is:
(a) Chlorine gas
(b) Calcium oxide

A
(a) $$6.022\times 10^{-23}$$ g (b) $$9.29\times 10^{-21}$$ g

B
(a) $$11.78\times 10^{-23}$$ g (b) $$9.29\times 10^{-23}$$ g

C
(a) $$23.96\times 10^{-21}$$ g (b) $$9.29\times 10^{-23}$$ g

D
(a) $$11.78\times 10^{-21}$$ g (b) $$9.29\times 10^{-21}$$ g

Solution

Number of molecules $$=N_m$$ $$=100$$

Weight of 100 molecules $$=m$$

We know, $$N_m=\dfrac{m}{MW}\times 6.023\times 10^{23}$$

$$\Rightarrow {m}=\dfrac{N_m\times MW}{6.023\times 10^{23}} $$ .....(1)

Let us apply the equation (1) to each of the given cases -

(a) Chlorine gas (MW= $$71$$ g/mol)

$$m \ = \dfrac{100\times 71}{6.023\times 10^{23}}$$

$$=11.78\times 10^{-21}$$ g

(b) Calcium oxide (MW= $$56$$ g/mol)

$$m=\dfrac{100\times 56}{6.023\times 10^{23}}$$

$$=9.29\times 10^{-21}$$ g
Option $$D$$ is correct.
Question 8
1 / -0

Total number of atoms in 20 g of calcium carbonate is ________.

A
$$N_A$$

B
$$ 5N_A$$

C
$$ \dfrac {1}{5}N_A$$

D
none of the above

Solution

number of moles in $$20$$ g of $$CaCO_3$$

$$1$$ molecule has $$5$$ atoms.

$$20$$ of calcium carbonate has $$\Rightarrow \dfrac{20}{100}\times 5 \times N_A$$

= $$N_A$$

Hence, option $$A$$ is the answer.
Question 9
1 / -0

Avogadro's number is ?

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

B
$$6.066 \times 10^{-19}$$

C
$$6.032 \times 10^{24}$$

D
$$6.203 \times 10^{18}$$
Solution
Correct option: $$A$$
Explanation:
- Avagadro’s number is a number of particles(electrons, protons, etc.) found in a mole of a substance.
- Conversion between moles and grams depends on Avogadro's number.
- Avogadro's number has value $$6.022 \times10^{23}$$.
- For example: $$1$$ mole of $$C = $$ $$12 \ g = $$ $$6.022 \times 10^{23}$$ molecules
Question 10
1 / -0

The ratio by mass of sulphur and oxygen in $$SO_2$$ is:

A
1 : 2

B
2 : 1

C
1 : 1

D
1 : 4

Solution

In $$SO_{2}$$, molar mass of S = $$32\ g\ mole^{-1}$$ & molar mass of $$O_{2}$$ = $$32\ g\ mol^{-1}$$.

Therefore the ratio of their masses = 32:32 = 1:1.

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Atoms and Molecules Test - 29

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Atoms and Molecules Test - 29

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Question 1

$$360$$

$$240$$

$$118$$

$$310$$

Solution

Question 2

Mixtures

Compounds

Elements

Colloids

Solution

Question 3

Avagadro number

Mass number

Both (a) and (b) above

Density

Solution

Question 4

6.50 g

13 g.

26 g.

15 g.

Question 5

$$K_2OH$$ $$ = 95 g mol^{1}$$

$$ KOH = 56 g mol^{1}$$

$$K(OH)_2$$ $$ = 73 g mol^{1}$$

$$ KOH = 58 g mol^{1}$$

Solution

Question 6

80

90

100

110

Solution

Question 7

(a) $$6.022\times 10^{-23}$$ g (b) $$9.29\times 10^{-21}$$ g

(a) $$11.78\times 10^{-23}$$ g (b) $$9.29\times 10^{-23}$$ g

(a) $$23.96\times 10^{-21}$$ g (b) $$9.29\times 10^{-23}$$ g

(a) $$11.78\times 10^{-21}$$ g (b) $$9.29\times 10^{-21}$$ g

Solution

Question 8

$$N_A$$

$$ 5N_A$$

$$ \dfrac {1}{5}N_A$$

none of the above

Solution

Question 9

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

$$6.066 \times 10^{-19}$$

$$6.032 \times 10^{24}$$

$$6.203 \times 10^{18}$$

Solution

Question 10

1 : 2

2 : 1

1 : 1

1 : 4

Solution

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