Atoms and Molecules Test

Result

Atoms and Molecules Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The number of atoms present in $$10$$ gms of $$CaCO_{3}$$ is:

A
$$5N_{A}$$

B
$$0.5N_{A}$$

C
$$5$$

D
$$N_{A}$$

Solution

The molecular weight of calcium carbonate is $$100$$ g.

$$10$$ g of calcium carbonate corresponds to $$0.1$$ mole.

One molecule of calcuim carbonate contains $$5$$ atoms. Hence, $$0.1$$ mole of calcium carbonate will contain $$0.5N_A$$ atoms. Here, $$N_A$$ is the Avogadro number.
Question 2
1 / -0

Assertion: 1 mole $$O_{3} (ozone)=N$$ molecule of $$O_{3} (ozone)=3N$$ atoms of $$O (oxygen)=48g$$

Reason: A mole is the amount of matter that contains as many as objects as the number of atoms exactly in 12 g C-12 ($$N$$ = Avogadro number).

A
Assertion is correct but Reason is incorrect.

B
Assertion is incorrect but Reason is correct.

C
Both Assertion and Reason are correct and Reason is correct explanation of Assertion

D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion

Solution

$$1$$ mole of ozone weighs $$48$$ g (molecular weight). It contains Avogadro number ($$N$$) of molecules. Each ozone molecule contains $$3$$ oxygen atoms. Hence, $$1$$ mole of ozone contains $$3N$$ atoms of oxygen.
Question 3
1 / -0

Mass of one atom of oxygen is:

A
$$\frac {16}{6.023 \times 10^{23}}g$$

B
$$ \frac {32}{6.023 \times 10^{23}}g$$

C
$$ \frac {1}{6.023 \times 10^{23}}g$$

D
$$8\ u$$

Solution

Mass of $$6.023 \times 10^{23}$$ atoms of oxygen = Atomic mass of oxygen = 16 g

Mass of one atom of oxygen =$$\dfrac{16}{6.023\times 10^{23}}$$g
Question 4
1 / -0

Cation and anion present (if any) in $$CH_3COONa$$ are :

A
Cation; $$CH_3 COO^-$$ Anion; $$Na^+$$

B
Anion; $$CH_3 COO^-$$ Cation; $$Na^+$$

C
Cation; $$CH_3 COO^{2-}$$ Anion; $$Na^+$$

D
Anion; $$CH_3 COO^{2-}$$ Cation; $$Na^+$$

Solution

The cation and anion is $$CH_3COONa$$ are
Cation$$\rightarrow Na^+$$
Anion $$\rightarrow CH_3COO^-$$
Question 5
1 / -0

Which of the following statements is not true about an atom?

A
Atoms are not able to exist independently.

B
Atoms are the basic units from which molecules and ions are formed.

C
Atoms are always neutral in nature.

D
Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

Solution

Option A is false. Atoms are not able to exist independently. All the elements on the far right side of the periodic table (inert or noble gas) atoms can exist independently because they all fulfil the octet rule. They do not need bond to become stable.

Option B is true. Atoms are the basic units from which molecules and ions are formed. A molecule is comprised of two or more chemically bonded atoms. The atoms may be of the same type of element, or they may be different. Atoms can readily gain or lose electrons. If electrons are lost or gained by a neutral atom, a charged particle is formed called as an ion.

Option C is true. Atoms are always neutral in nature. The nucleus of an atom (contains protons and neutrons) remains unchanged after ordinary chemical reactions, but atoms can readily gain or lose electrons.

Option D is true. Atoms aggregate in large numbers to form the matter that we can see, feel or touch. Atoms and molecules are all composed of matter. Matter is anything that has mass and takes up space. Matter exists in 3 states gases, liquid and solid.
Question 6
1 / -0

Which of the following compounds contains maximum number of atoms?

A
18gm of $$H_2O$$

B
18gm of $$O_2$$

C
18gm of $$CO_2$$

D
18gm of $$CH_4$$

Solution

$$No. of\ atoms= \dfrac{Mass \times atomicity}{Molar\ Mass}\times N_{A}$$

Using the above formula:
a) In 18gm of $$ H_{2}O$$:
No. of atoms= $$\dfrac{18\times 3}{18}\times N_{A} = 3N_{A}$$

b) In 18gm of $$ O_{2}$$:
No of atoms= $$\dfrac{18\times 2}{32}\times N_{A} = 1.12N_{A}$$

c) In 18gm of $$ CO_{2}$$:
No. of atoms= $$\dfrac{18\times 3}{44}\times N_{A} = 1.23N_{A}$$

d) In 18gm of $$ CH_{4}$$:
No. of atoms= $$\dfrac{18\times 5}{16}\times N_{A} = 5.63N_{A}$$
Therefore, 18gm $$CH_{4}$$ has a maximum number of atoms.
Question 7
1 / -0

The number of atoms in 0.1 mol of a triatomic gas is:

A
6.026 x 10$$^{22}$$

B
1.806 x 10$$^{23}$$

C
3.600 x 10$$^{23}$$

D
1.800 x 10$$^{22}$$

Solution

A number of atoms $$=$$ No. of moles x Avagadro's Number
$$=0.1 \times 6.022\times 10^{23}$$
$$= 6.02\times 10^{22}$$
Now since we have the number of molecules of the gas. And also we know that there are 3 atoms present in one molecule.
So the number of atoms present in 0.1 moles of the gas is:
$$=3×6.02×10^{22}$$
$$=18.06×10^{22}$$
$$=1.806×10^{23}$$
Question 8
1 / -0

Which of the following would weigh the highest?

A
0.2 mole of sucrose $$(C_{12} H_{22} O_{11})$$

B
2 moles of carbon dioxide $$(CO_2)$$

C
2 moles of calcium carbonate $$(CaCO_3)$$

D
10 moles of water $$(H_2O)$$

Solution

Applying mole concept,
1 mole $$=$$ molecular weight in gram.
Molecular weight of $$C_{12}H_{22}O_{11} = 12\times12 + 22\times 1 + 11\times 16 = 342$$
Molecular weight of $$CO_2 = 12 + 2\times 16 = 44$$
Molecular weight of $$CaCO_3 = 40 + 12 + 3\times 16 = 100$$
Molecular weight of $$H_2O = 16 + 2\times 1 = 18$$

The results can be calculated as:
Weight of a sample in gram $$=$$ number of moles $$\times$$ molar mass
(a) 0.2 moles of $$C_{12}H_{22}O_{11} = 0.2 \times 342 = 68.4\ gm$$
(b) 2 moles of $$CO_2 = 2 \times 44 = 88\ gm$$
(c) 2 moles of $$CaCO_3 = 2 \times 100 = 200\ gm$$
(d) 10 moles of $$H_2O = 10 \times 18 = 180\ gm$$

Hence, 2 moles of $$CaCO_3$$ weighs more than other compounds.
Question 9
1 / -0

Dalton's atomic theory successfully explained :
(i) Law of conservation of mass
(ii) Law of constant composition
(iii) Law of radioactivity
(iv) Law of multiple proportions

A
(i), (ii) and (iii)

B
(i), (iii) and (iv)

C
(ii), (iii) and (iv)

D
(i), (ii) and (iv)
Solution
Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of different sizes and masses.

Dalton's atomic theory explains:
Law of conservation of mass
Law of constant composition
Law of multiple proportions
Question 10
1 / -0

The mass of one steel screw is 4.11gm. Find the mass of one mole of these steel screws.

A
$$2.475 \times 10^{21} kg$$

B
$$3.56 \times 10^{21} kg$$

C
$$4.2 \times 10^{21} kg$$

D
$$4.2 \times 10^{20} kg$$

Solution

Mass of one steel screw is 4.11 gm, so mass of one mole is $$ = 4.11*6.023*10^{23} = 2.475 \times 10^{24} g = 2.475 \times 10^{21} kg$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Atoms and Molecules Test - 27

Result

Atoms and Molecules Test - 27

Score

-

Rank

-

SHARING IS CARING

Question 1

$$5N_{A}$$

$$0.5N_{A}$$

$$5$$

$$N_{A}$$

Solution

Question 2

Assertion is correct but Reason is incorrect.

Assertion is incorrect but Reason is correct.

Both Assertion and Reason are correct and Reason is correct explanation of Assertion

Both Assertion and Reason are correct but Reason is not correct explanation of Assertion

Solution

Question 3

$$\frac {16}{6.023 \times 10^{23}}g$$

$$ \frac {32}{6.023 \times 10^{23}}g$$

$$ \frac {1}{6.023 \times 10^{23}}g$$

$$8\ u$$

Solution

Question 4

Cation; $$CH_3 COO^-$$ Anion; $$Na^+$$

Anion; $$CH_3 COO^-$$ Cation; $$Na^+$$

Cation; $$CH_3 COO^{2-}$$ Anion; $$Na^+$$

Anion; $$CH_3 COO^{2-}$$ Cation; $$Na^+$$

Solution

Question 5

Atoms are not able to exist independently.

Atoms are the basic units from which molecules and ions are formed.

Atoms are always neutral in nature.

Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

Solution

Question 6

18gm of $$H_2O$$

18gm of $$O_2$$

18gm of $$CO_2$$

18gm of $$CH_4$$

Solution

Question 7

6.026 x 10$$^{22}$$

1.806 x 10$$^{23}$$

3.600 x 10$$^{23}$$

1.800 x 10$$^{22}$$

Solution

Question 8

0.2 mole of sucrose $$(C_{12} H_{22} O_{11})$$

2 moles of carbon dioxide $$(CO_2)$$

2 moles of calcium carbonate $$(CaCO_3)$$

10 moles of water $$(H_2O)$$

Solution

Question 9

(i), (ii) and (iii)

(i), (iii) and (iv)

(ii), (iii) and (iv)

(i), (ii) and (iv)

Solution

Question 10

$$2.475 \times 10^{21} kg$$

$$3.56 \times 10^{21} kg$$

$$4.2 \times 10^{21} kg$$

$$4.2 \times 10^{20} kg$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.