Atoms and Molecules Test

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Atoms and Molecules Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

Irrespective of the source, a pure sample of water always yields $$88.89\%$$ oxygen by mass and $$11.11\%$$ hydrogen by mass. This is explained by the law of:

A
conservation of mass

B
constant proportion

C
multiple proportion

D
none of these

Solution

The molecular mass of water is 18 amu.

Molecular mass of oxygen is 16 amu. So, % of oxygen in $$H_2O$$ is $$ \dfrac{16}{18}\times 100= 88.89 \%.$$.

Molecular mass of hydrogen is 2 amu. So, % of hydrogen in $$H_2O$$ is$$ \dfrac{2}{18}\times 100= 11.11\%.$$.

According to the law of constant proportions, a chemical compound always contains the same elements combined in the same proportion by mass. For example, pure water obtained from different sources such as rivers, wells, springs, seas, etc., always contains hydrogen and oxygen together in the ratio of 1:8 by mass.
Question 2
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$$8$$ g of $$O_{2}$$ has the same number of molecules as:

A
7 gms of $$CO$$

B
14 gms of $$CO$$

C
22 gms of $$CO_{2}$$

D
44 gms of $$CO_{2}$$

Solution

Moles of $$O_2 = \dfrac{8}{32} = 0.25$$ moles

$$No.\ of\ molecules$$ $$= No.\ of\ moles \times N_{A}$$.

Thus, no. of moles is directly proportional to no. of molecules.

Moles of $$7$$ gm $$CO = \dfrac{7}{28} = 0.25$$

Moles of $$14$$ gm $$CO = \dfrac{14}{28} = 0.5$$

Moles of $$22$$ gm $$CO_{2} = \dfrac{22}{44} = 0.5$$

Moles of $$44$$ gm of $$CO_{2} = 1$$

Hence, option A is correct.
Question 3
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Choose the correct option for the following statement.

The use of $$^{12}C$$ scale has superseded the older scale of atomic mass based on $$^{16}O$$ isotope, one important advantage of the former being:

A
The atomic masses on $$^{12}C$$ scale became whole numbers.

B
$$^{12}C$$ is more abundant in the earths crust than $$^{16}O$$.

C
The difference between the physical and chemical atomic masses got narrowed down significantly.

D
$$^{12}C$$ is situated midway between metals and non-metals in the periodic table.

Solution

Physical stuck to pure $$^{16}O$$ for mass scale and chemistry used an average of all three isotopes because natural oxygen is the mixture of $$^{16}O$$, $$^{17}O$$ and $$^{18}O$$. There was not much difference on average.
But with molecules of 200-500 amu, it starts to add up. Use of $$^{12}C$$ which is isotopically pure, get narrow down the difference between physical and chemical atomic masses.
Question 4
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The number of molecules present in $$4.4$$ g of $$CO_{2}$$ gas is:

A
$$6.022\times 10^{23}$$

B
$$5.022\times 10^{23}$$

C
$$6.022\times 10^{24}$$

D
$$6.022\times 10^{22}$$

Solution

1 mole of $$CO_2$$ have $$6.022\times10^{23}$$ molecules.

Weight of one 1 mole of $$CO_2$$ = 44g(molar mass).

Number of moles = $$\dfrac {given \ mass}{molar \ mass}$$.

Number of moles of $$CO_2$$ in 4.4 g = $$\dfrac{4.4}{44}=0.1$$.

So, number of molecules in 4.4g of $$CO_2$$ = $$\dfrac{6.022\times10^{23}}{44}\times4.4 = 6.023 \times10^{22}$$.
Question 5
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Which of the following is the best example to demonstrate the law of conservation of mass?

A
12 gm of carbon combines with 32 gm of oxygen to form 44 gm of $$CO_{2}$$.

B
When 72 gm of carbon is heated in a vacuum and no change in its mass takes place.

C
The weight of a piece of platinum is the same before and after heating in air.

D
None of these

Solution

According to the law of conservation of mass:
The total mass of the reactant = the total mass of the product.

Mass of reactants $$= 12 + 32 = 44$$ i.e w(C)+ w(oxygen)

Mass of products $$= 44$$ i.e. w(carbon dioxide)

Both B and C do not represent a chemical reaction, but only heating of elements, hence they are not the best examples to demonstrate the law of conservation of mass.
Question 6
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The weight of gaseous mixture containing $$6.02 \times 10^{23}$$ molecules of nitrogen and $$3.01\times 10^{23}$$ molecules of sulphur dioxide is:

A
46

B
92

C
60

D
30

Solution

The molar masses of nitrogen and sulphur dioxide are $$28$$ g/mol and $$32$$ g/mol respectively.
$$6.02 \times 10^{23}$$ molecules of nitrogen corresponds to $$1$$ mole or $$28$$ g.
$$3.01\times 10^{23}$$ molecules of sulphur dioxide corresponds to $$0.5$$ mole or $$32$$ g.
Hence, the weight of the gaseous mixture is $$28 \: g + 32 \: g = 60 \: g$$.
Question 7
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Number of atoms in $$558.5$$ gram Fe is:
[Atomic mass of $$Fe = 55.85\ u$$]

A
twice that of 60 g carbon

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

C
half that in $$8$$g He

D
$$558.5\times6.023\times 10^{23}$$

Solution

Number of moles of Fe atoms = $$\dfrac {Mass}{Atomic\ mass}$$ $$=\dfrac {558.5}{55.85} = 10$$

558.5 g of Fe corresponds to 10 moles of Fe and has $$ 10N_{A}$$ atoms of Fe.

Option A - 12 g of carbon means Avogadro's number, therefore, 60 g of carbon means, 5 moles of carbon and thus contains $$ 5\times N_{A}$$ atoms, twice of this equals $$10 N_{A}$$ atoms.

Option B - $$ \dfrac{N_{A}} {10}$$ atoms.

Option C - 8 g of He corresponds to 2 moles of He and thus, contains $$ 2N_{A}$$ atoms, half of the atoms will be $$N_{A}$$ atoms of He.

So, option A only corresponds to number of atoms in 558.5 g Fe, i.e. $$10N_A$$ atoms.
Question 8
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$$10$$ grams each of $$O_{2},\ N_{2}$$ and $$Cl_{2}$$ are kept in three bottles. The correct order of arrangement of bottles containing decreasing number of molecules is:

A
$$O_{2} >\ N_{2}>\ Cl_{2}$$

B
$$Cl_{2}>\ N_{2}>\ O_{2}$$

C
$$Cl_{2}>\ O_{2}>\ N_{2}$$

D
$$N_{2}>\ O_{2}>\ Cl_{2}$$

Solution

No. of molecules $$=\displaystyle \dfrac{Weight}{Molecular\ Weight}\times N_A$$

Molecular weight order is $$Cl_2 > O_2 > N_2$$.

Therefore, the order of no. of molecules is $$N_{2}>O_{2}>Cl_{2}$$.

Hence, option $$D$$ is correct.
Question 9
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Which of the following has highest mass?

A
$$50$$ g of iron

B
$$5$$ moles of nitrogen gas

C
$$5 \times 10^{23}$$ atoms of carbon

D
All are equal

Solution

$$Number\ of\ moles = \dfrac {Mass}{Molar\ mass}$$

$$\therefore Mass = Number\ of\ moles\ \times molar\ mass$$

To calculate the mass of 5 moles of nitrogen gas, we have first find out its molar mass.
So, molar mass of $$N_2 = 2 \times 14 = 28\ g/mol$$
$$Mass\ of\ 5\ moles\ of\ N_2 = Number\ of\ moles\ \times molar\ mass$$ $$= 5 \times 28 = 140\ g$$

Option B (140 g) is greater than option A (50 g).

Let's now compare with option C ($$5 \times 10^{23}$$ atoms of carbon)

$$Number\ of\ moles = \dfrac {Given\ number\ of\ particles}{Avogadro\ number}$$

$$Mass = Number\ of\ moles\ \times molar\ mass$$

$$\therefore Mass = \dfrac {Given\ number\ of\ particles}{Avogadro\ number} \times molar\ mass$$ $$= \dfrac {5 \times 10^{23}}{ 6.022 \times 10^{23}} \times 12 = 19.96\ g$$

So, mass of option B is higher than both option A and option C.
Question 10
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Which one of the following contains a same number of atoms as there are in 12 grams of magnesium?

A
12 grams of carbon

B
40 grams of calcium

C
16 grams of oxygen

D
7 grams of carbon monoxide

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Atoms and Molecules Test - 25

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Atoms and Molecules Test - 25

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Question 1

conservation of mass

constant proportion

multiple proportion

none of these

Solution

Question 2

7 gms of $$CO$$

14 gms of $$CO$$

22 gms of $$CO_{2}$$

44 gms of $$CO_{2}$$

Solution

Question 3

The atomic masses on $$^{12}C$$ scale became whole numbers.

$$^{12}C$$ is more abundant in the earths crust than $$^{16}O$$.

The difference between the physical and chemical atomic masses got narrowed down significantly.

$$^{12}C$$ is situated midway between metals and non-metals in the periodic table.

Solution

Question 4

$$6.022\times 10^{23}$$

$$5.022\times 10^{23}$$

$$6.022\times 10^{24}$$

$$6.022\times 10^{22}$$

Solution

Question 5

12 gm of carbon combines with 32 gm of oxygen to form 44 gm of $$CO_{2}$$.

When 72 gm of carbon is heated in a vacuum and no change in its mass takes place.

The weight of a piece of platinum is the same before and after heating in air.

None of these

Solution

Question 6

46

92

60

30

Solution

Question 7

twice that of 60 g carbon

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

half that in $$8$$g He

$$558.5\times6.023\times 10^{23}$$

Solution

Question 8

$$O_{2} >\ N_{2}>\ Cl_{2}$$

$$Cl_{2}>\ N_{2}>\ O_{2}$$

$$Cl_{2}>\ O_{2}>\ N_{2}$$

$$N_{2}>\ O_{2}>\ Cl_{2}$$

Solution

Question 9

$$50$$ g of iron

$$5$$ moles of nitrogen gas

$$5 \times 10^{23}$$ atoms of carbon

All are equal

Solution

Question 10

12 grams of carbon

40 grams of calcium

16 grams of oxygen

7 grams of carbon monoxide

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