Atoms and Molecules Test

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Atoms and Molecules Test - 26

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Weekly Quiz Competition

Question 1
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Which of the following gases contain the same number of molecules as that of $$16$$ g of oxygen?

A
$$16$$ g of $$O_{3}$$

B
$$32$$ g of $$SO_{2}$$

C
$$16$$ g of $$SO_{2}$$

D
None of these

Solution

Number of molecules = Number of moles$$\times N_A$$

When two gases have the same number of molecules, they will contain the same number of moles.

Number of moles of a substance= $$\dfrac{give\ mass}{molar\ mass}$$

$$16$$ g of oxygen $$(O_2$$) corresponds to $$\dfrac{16}{32} = 0.5$$ mole.

$$16$$ g of $$O_3$$ corresponds to $$\dfrac{16}{48} = 0.33$$ mole.

$$32$$ g of $$SO_2$$ corresponds to $$\dfrac{32}{64} = 0.5$$ mole.

$$16$$ g of $$SO_2$$ corresponds to $$\dfrac{16}{64} = 0.25$$ mole.

Hence, option B is correct.
Question 2
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$$4.6\times 10^{22}$$ atoms of an element weight $$13.8$$ g. What is the atomic mass of the element?

A
290 u

B
180.6 u

C
34.4 u

D
104 u

Solution

$$1$$ mole of any substance contains $$6.02 \times 10^{23}$$ atoms.

Thus, $$4.6 \times 10^{22}$$ atoms corresponds to $$\dfrac{4.6 \times 10^{22}}{6.022 \times 10 ^{23}}=0.0764$$ moles.

$$0.0764$$ moles weighs $$13.8$$ g.

Thus, $$1$$ mole will weigh $$\dfrac{13.8}{0.0764}= 180.6 \ g$$.

Hence, the atomic mass of the element will be 180.6 u.
Question 3
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The ratio of the mass of a C-12 atom to that of an atom of element X (whose atomicity is four) is $$1: 9$$. The molecular mass of element X is :

A
$$480\ gmol^{-1}$$

B
$$432\ gmol^{-1}$$

C
$$36\ gmol^{-1}$$

D
$$84\ gmol^{-1}$$

Solution

The mass of one mole of carbon-12 is 12 g.
The mass of one mole of X atoms will be $$9 \times 12 = 108$$ g.
But here atomicity of $$X$$ is 4.
Thus, one molecule will contain $$4X$$ atoms.
Hence, the mass of one mole of the compound will be $$4 \times 108 = 432$$ g. This is the molecular weight.
Hence, option $$B$$ is correct.
Question 4
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The weight of one molecule of $$C_{60} H_{122}$$ is:

A
$$1.2\times 10^{-20}g$$

B
$$5.025\times 10^{23}g$$

C
$$1.4\times 10^{-21}g$$

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

Solution

The molecular weight of $$C_{60}H_{122}$$ is $$842$$ g/ mol.

One mole of $$C_{60}H_{122}$$ weighs $$6.02 \times 10^{23}$$ g.

Thus, weight of one molecule of $$C_{60}H_{122}$$ is $$\dfrac { 842 }{ 6.02 \times 10^{ 23 } } = 1.4 \times 10^{-21} g$$.

Hence, option $$C$$ is correct.
Question 5
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Which of these has the maximum number of molecules?

A
$$7$$ g $$N_{2}$$

B
$$2$$ g $$H_{2}$$

C
$$16$$ g $$NO_{2}$$

D
$$16$$ g $$O_{2}$$

Solution

$$\bf{Hint-}$$ $$1$$ mole of any compound containd $$N_A$$ molecules.

$$\bf{Formula\ used-}$$
The expression for the number of molecules is given as:

$$Number\: of\: molecules=\dfrac { Mass}{ Molecular\: mass } \times N_A $$,

where $$N_A$$ is the Avogadro number.

$$\bf{Explanation-}$$

$$\bullet$$ $$N_2 \: contains\: \dfrac { 7\times N_A }{ 28 } =0.25 \times N_A\: molecules$$

$$\bullet$$ $$H_{ 2 }\: contains\: \dfrac { 2\times N_A }{ 2 } =1 \times\ N_A\: molecules$$

$$\bullet$$ $$NO_{ 2 }\: contains\: \dfrac { 16\times N_A }{ 46 } =0.35 \times N_A\: molecules$$

$$\bullet$$ $$O_{ 2 }\: contains\: \dfrac { 16\times N_A }{ 32 } = 0.5 \times N_A\:molecules$$

$$\bf{Final\ answer-}$$ $$B$$
Question 6
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Which of the following contains highest number of atoms?

A
4 gm of oxygen

B
4 gm of carbon dioxide

C
4 gm of helium

D
4 gm of methane

Solution

No. of molecules $$= No.\ of\ moles \times \ N_A$$ $$= \dfrac {Mass}{Molar\ mass} \times \ N_A$$

No. of atoms $$= \dfrac {Mass}{Molar\ mass} \times \ N_A \times No.\ of\ atoms\ in\ 1\ molecule$$

Option A: Number of atoms in 4 g oxygen $$O_2$$ $$=\dfrac{4}{32} \times N_A \times 2 = \dfrac{N_A} {4}$$

Option B: Number of atoms in 4 g carbon dioxide $$CO_2$$ $$ =\dfrac{4 }{44} \times N_A\times 3 = \dfrac{N_A} {3.67}$$

Option C: Number of atoms in 4 g helium $$He$$ $$ =\dfrac{4}{4} \times N_A \times 1= \dfrac{N_A} {1}$$

Option D: Number of atoms in 4 g methane $$CH_4$$ $$ =\dfrac{4 }{16} \times N_A \times 5 = \dfrac{N_A} {0.8}$$

Arranging in the order of increasing number of atoms:
$$\dfrac{N_A} {4}$$ < $$ \dfrac{N_A} {3.67}$$ < $$ \dfrac{N_A} {1}$$ < $$\dfrac{N_A} {0.8}$$

Hence option $$D$$ is the correct option having highest number of atoms.
Question 7
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Different samples of water were found to contain hydrogen and oxygen in the ratio of 1 : 8 by mass. This proves the law of:

A
multiple proportion

B
constant proportion

C
reciprocal proportion

D
conservation of mass

Solution

The Law of Constant proportion states that "A chemical compound always contains the same elements combined in the same proportion by mass."
Question 8
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The maximum number of molecules are present in:

A
$$36$$ g $$H_{2} O$$

B
$$28$$ g $$CO$$

C
$$46$$ g $$C_{2} H_{5} OH$$

D
$$54$$ g $$N_{2} O_{5}$$

Solution

Number of molecules = Number of moles $$\times N_A$$

$$= \dfrac {\text{given mass}} {\text{molar mass}}\times N_A$$

The more the number of moles, the more will be the number of molecules.

$$36$$ g of water($$H_2O$$) corresponds to $$\dfrac{36}{18} = 2$$ moles (Molar mass 18 $$ g\ mol^{-1}$$).

$$28$$ g of $$\ CO$$ corresponds to $$\dfrac{28}{28} = 1$$ mole (Molar mass 28 $$ g\ mol^{-1}$$).

$$46$$ g ethanol($$C_2H_5OH$$) corresponds to $$\dfrac{46}{46} = 1$$ mole (Molar mass 46 $$ g\ mol^{-1}$$).

$$54$$ g dinitrogen pentoxide($$N_2O_5$$) corresponds to $$\dfrac{54}{108} = 0.5$$ moles (Molar mass 108 $$g\ mol^{-1}$$).

Hence, option A is correct.
Question 9
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The mass of one oxygen molecule is:

A
$$2.656\times 10^{-23}\ gm$$

B
$$5.312\times 10^{-23}\ gm$$

C
$$1.66\times 10^{24}\ gm$$

D
$$32\ gm$$

Solution

$$\bf{Hint-}$$ One mole of any gas has Avogadro's number of molecules.

$$\bf{STEP-1}$$ To find the mass of one-mole oxygen gas molecules.

One mole contains $$6.02\times 10^{ 23 } \ molecules$$ of oxygen.

$$1\ mole$$ of oxygen weighs $$32\ g$$.

Thus, weight of $$1$$ oxygen molecule is $$ \dfrac { 32 }{ 6.023\times 10^{ 23 } } =5.312\times10^{ -23 }\ gm$$.

$$\bf{Final answer-}$$ Option B
Question 10
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Which of the following contain maximum number of molecules?

A
1g $$CO_2$$

B
1g $$N_2$$

C
1g $$H_2$$

D
1g $$CH_4$$

Solution

For a molecule,
molecular weight in grams $$=$$ Avogadro's number$$({ N }_{ A }) = 6.023 \times { 10 }^{ 23 }$$
(A) 1g $$CO_2$$
44 gram of  $${ CO }_{ 2 }$$ =$${ N }_{ A }$$ molecules
$$ \therefore$$ 1 gram of $$CO_2$$ contains $$ \dfrac{1}{44} \times N_A$$ molecules

(B) 1g $$ { N }_{ 2 }$$
28 gram of   $$ { N }_{ 2 } ={ N }_{ A }$$ molecules
$$\therefore $$1 gram of  $$ { N }_{ 2 }$$ contains $$\dfrac{1}{28} \times N_A $$ molecules

(C) 1g $$H_2$$
2 gram of $$H_2$$ contains $$N_A$$molecules
$$ \therefore$$ 1 gram of $$H_2$$ contains $$\dfrac {1}{2} \times N_A$$ molecules

(D) 1g  $${ CH }_{ 4 }$$
16 gram of $${ CH }_{ 4 }$$ =$${ N }_{ A }$$ molecules
$$\therefore$$ 1 gram of   $${ CH }_{ 4 }$$ =$${ N }_{ A }/16$$
Maximum number of molecules are found in 1 gram hydrogen.

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Atoms and Molecules Test - 26

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Atoms and Molecules Test - 26

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Question 1

$$16$$ g of $$O_{3}$$

$$32$$ g of $$SO_{2}$$

$$16$$ g of $$SO_{2}$$

None of these

Solution

Question 2

290 u

180.6 u

34.4 u

104 u

Solution

Question 3

$$480\ gmol^{-1}$$

$$432\ gmol^{-1}$$

$$36\ gmol^{-1}$$

$$84\ gmol^{-1}$$

Solution

Question 4

$$1.2\times 10^{-20}g$$

$$5.025\times 10^{23}g$$

$$1.4\times 10^{-21}g$$

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

Solution

Question 5

$$7$$ g $$N_{2}$$

$$2$$ g $$H_{2}$$

$$16$$ g $$NO_{2}$$

$$16$$ g $$O_{2}$$

Solution

Question 6

4 gm of oxygen

4 gm of carbon dioxide

4 gm of helium

4 gm of methane

Solution

Question 7

multiple proportion

constant proportion

reciprocal proportion

conservation of mass

Solution

Question 8

$$36$$ g $$H_{2} O$$

$$28$$ g $$CO$$

$$46$$ g $$C_{2} H_{5} OH$$

$$54$$ g $$N_{2} O_{5}$$

Solution

Question 9

$$2.656\times 10^{-23}\ gm$$

$$5.312\times 10^{-23}\ gm$$

$$1.66\times 10^{24}\ gm$$

$$32\ gm$$

Solution

Question 10

1g $$CO_2$$

1g $$N_2$$

1g $$H_2$$

1g $$CH_4$$

Solution

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