Atoms and Molecules Test

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Atoms and Molecules Test - 69

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Weekly Quiz Competition

Question 1
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$$\displaystyle n_{1}$$ g of substance X reacts with $$\displaystyle n_{2}$$ g of substance Y to form $$\displaystyle m_{1}$$ g of substance R and $$\displaystyle m_{2}$$ g of substance of S. This reaction can be represented as $$X + Y \rightarrow R + S$$. The relation which can be established in the amounts of the reactants and the products will be :

A
$$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$$

B
$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

C
$$\displaystyle n_{1}=n_{2}$$

D
$$\displaystyle m_{1}=m_{2}$$

Solution

The relation which can be established in the amounts of the reactants and the products will be as follows:

$$\displaystyle \text {Total mass of reactants}=\text {Total mass of products}$$

$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

This is based on the law of conservation of mass.
Question 2
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The most accurate statement with regard to the significance of Avogadro's number, $$6.02 \times {10}^{23}$$.

A
It is the conversion factor between grams and atomic mass units

B
It is a universal physical constant just as the speed of particle

C
It is the number of particles that is required to fill a $$1$$-liter container

D
It is the inverse diameter of an $$H$$ atom

E
It is the number of electrons in the universe

Solution

Avogadro's number ($$6.022 \times 10^{23}$$) is the conversion factor between the grams and atomic mass units.
Example:
Atomic mass unit of $$O$$-atom $$= 16 u$$
Actual mass of $$O$$-atom $$ = \dfrac {16}{6.022 \times 10^{23}} = 2.7 \times 10^{-23} g$$
As $$6.022\times 10^{23}$$ O-atoms have $$16g$$ molar mass, which is the same numerical value for atomic mass but a different unit, i.e., $$amu$$ = atomic mass unit (amu).
Question 3
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Molecules are made up of ..........

A
electrons

B
protons

C
neutrons

D
atoms

E
none of these

Solution

An atom is made up of electrons, protons, and neutrons. A molecule is formed when atoms of the same or different elements combine. For example, the water molecule is formed when atoms of oxygen and hydrogen combine. Hence option d is correct.
Question 4
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A charged particle formed by the loss or gain of electrons during a compound formation is called an .....

A
ion

B
Cation

C
Anion

D
Positron

Solution
Question 5
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$$1.375$$ g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was $$1.098$$ g. In another experiment, $$1.156$$ g of copper was dissolved in nitric acid and the resulting solution was evaporated to dryness. The residue of copper nitrate when strongly heated was converted into $$1.4476$$ g of cupric oxide. State the law illustrated by these chemical combinations.

A
Law of reciprocal proportion

B
Law of multiple proportion

C
Law of constant composition

D
None of these

Solution

Law of constant composition states that all samples of a given chemical compound have the same elemental composition by mass.

Here, in the first experiment, 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was 1.098 g.

1.375 g of cupric oxide has 1.098 g of copper and $$1.375-1.098 = 0.277$$ g of oxygen.

Copper and oxygen are in the ration $$\cfrac{1.098}{0.277} = \cfrac{1}{4}$$.

In the second experiment, 1.156 g of copper was dissolved in nitric acid to form copper nitrate, which on strongly heating got converted into 1.4476 g of cupric oxide.

=1.4476 g of cupric oxide has 1.156 g of copper and $$ 1.4476-1.156 =0.2911 g $$ oxygen.

Copper and oxygen are in the ratio $$\cfrac{1.156}{0.2911} = \cfrac{1}{4}$$.

This means whatever may be the source of a compound, it has the same elemental composition by mass in all samples.
Question 6
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What is the ratio of the number of atoms of elements potassium, chlorine, and oxygen respectively present in $$245 g$$ of $$KCl{ { O }_{ 3 } }$$?

A
$$2 : 2 : 3$$

B
$$1 : 1 : 3$$

C
$$39 : 36 : 48$$

D
$$1 : 2 : 6$$

Solution

$$KClO_3\longrightarrow$$ molecular weight = $$122.5gmol^{-1}$$

There is $$1$$ mole of potassium in $$1$$ mole of $$KClO_3=1\times 39=39g$$
There is $$1$$ mole of $$Cl=1\times 35.5=35.5g$$
and $$3$$ mole of oxygen $$=3\times 16=48g$$

Moles of $$KClO_3=\dfrac {Given\quad weight}{molecular\quad weight}$$
$$=\cfrac {245g}{122.5gmol^{-1}}$$

Moles of $$KClO_3=2$$
$$\therefore$$ Moles of Potassium $$= 2$$
Moles of Chlorine $$= 2$$
Moles of Oxygen $$= 6$$

(1) Number of atoms of $$K$$ = Moles $$\times$$ Avogadro's number
$$=2\times 6.022\times 10^{23}$$
$$=12.044\times 10^{23}$$

(2) Number of atoms of $$Cl$$ = $$2\times 6.022\times 10^{23}$$
$$=12.044\times 10^{23}$$

(3) Number of atoms of Oxygen = $$6\times 6.022\times 10^{23}$$
$$=36.132\times 10^{23}$$

Ratio of number of atoms of $$K,Cl$$ & $$O=1:1:3$$ .
Question 7
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On heating $$KClO_{3}$$ at a certain temperature, it is observed that one mole of $$KClO_{3}$$ yields one mole of $$O_{2}$$. What is the mole fraction of $$KClO_{4}$$ in the final solid mixture containing only $$KCl$$ and $$KClO_{4}$$, the latter being formed by the parallel reaction?

A
$$0.50$$

B
$$0.25$$

C
$$0.75$$

D
$$0.67$$

Solution

Given,
$$1$$ $$KClO_3$$ decomposes to give $$1$$ mole $$O_2$$
The reaction involved is
$$2KClO_3\longrightarrow 2KCl+3O_2$$
It is clear from above equation that $$2$$ moles $$KClO_3$$ produce $$3$$ moles $$O_2$$ but it is given that $$1$$ mole $$O_2$$ is produced.
$$\Rightarrow$$ Moles of $$KClO_3$$ used= $$\cfrac {1\times 2}{3}=0.66$$ moles$$\longrightarrow (1)$$
$$\therefore$$ The left over moles of $$KClO_3$$ are $$1-0.66=0.34$$ moles $$\longrightarrow (2)$$
Now the parallel reaction is
$$4KClO_3\longrightarrow 3KClO_4+KCl$$
It is clear that $$4$$ moles $$KClO_3$$ gives $$3$$ moles $$KClO_4$$ & $$1$$ mole $$KCl \longrightarrow (3)$$
Now, for determining mole fraction of $$KClO_4$$ first calculate moles of $$KClO_4$$ & $$KCl$$
From $$(2)$$ & $$(3)$$
If $$4$$ moles $$KClO_3$$ gives $$3$$ moles $$KClO_4$$
$$\Rightarrow 0.34$$ moles $$KClO_3$$ give $$\cfrac {0.34\times 3}{4}$$
$$\therefore$$ No. of moles of $$KClO_4$$ produced=$$0.255$$ moles
Now, if $$4$$ moles $$KClO_3$$ give $$1$$ mole $$KCl$$
$$\Rightarrow 0.34$$ moles $$KClO_3$$ give $$\cfrac {0.34\times 1}{4}$$ moles
$$\therefore$$ No. of Moles of $$KCl$$ produced=$$0.085$$ moles
$$\therefore$$ Total no. of Moles in final mix=$$0.255+0.085$$
$$=0.34$$ moles
$$\therefore$$ Mole fraction of $$KClO_4=\cfrac {Moles\quad of\quad KClO_4}{Total\quad moles}$$
$$=\cfrac {0.255}{0.34}$$
$$=0.75$$
Question 8
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Which of the following contains the greatest number of atoms ?

A
1.0 g of butane $$(C_4H_{10})$$

B
1.0 g of nitrogen $$(N_2)$$

C
1.0 g of silver ($$Ag$$)

D
1.0 g of water $$(H_2O)$$

Solution
Question 9
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The atomic weight of silicon is $$28.086$$ and that of oxygen is $$15.9994$$ (on the $$C^{12}$$ scale). Calculate the mass in grams of a single molecule of $$SiO_2$$.

A
$$9.977\times 10^{-23}gm$$

B
$$9.8\times 10^{-22}gm$$

C
$$1\times 10^{-22}gm$$

D
$$6.6\times 10^{-23}gm$$

Solution

Molecular weight of $$SiO_{2}$$ is $$ 28.086 + 2\times 15.9994= 60.08$$ u

$$\therefore$$ 1 mole of $$SiO_{2}$$ weighs 60.08 g

$$\Rightarrow$$ $$6.022 \times 10^{23} $$ moelcules weigh 60.08 g of $$ SiO_{2} $$ (By Avogadro's Rule)

1 molecule of $$ SiO_{2} $$ contains $$\dfrac {60.08g}{6.022 \times 10^{23} } \times 1 = 9.977 \times 10^{-23} g $$

$$\therefore$$ Mass of a single molecule of $$ SiO_{2} = 9.977 \times 10^{-23}g.$$
Question 10
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The correct increasing order of molecular masses is:

A
$$H_{2}O > H_{2}S > CO_{2} > SO_{2}$$

B
$$H_{2}O > H_{2}S < CO_{2} > SO_{2}$$

C
$$H_{2}O < H_{2}S < CO_{2} < SO_{2}$$

D
$$H_{2}O > H_{2}S > CO_{2} < SO_{2}$$

Solution

(i) $$H_2O \Rightarrow 2+16 = 18 u$$

(ii) $$H_2S$$ $$\Rightarrow 2\times 1+1\times 32=2+32=34 u$$

(iii) $$CO_2$$ $$\Rightarrow 1\times 12+2\times 16=12+32=44 u$$

(iv) $$SO_2$$ $$\Rightarrow 1\times 32+2\times 16=32+32=64 u$$

Hence, $$H_2O<H_2S<CO_2<SO_2$$ is the correct order of increasing molecular mass.

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Atoms and Molecules Test - 69

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Question 1

$$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$$

$$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$$

$$\displaystyle n_{1}=n_{2}$$

$$\displaystyle m_{1}=m_{2}$$

Solution

Question 2

It is the conversion factor between grams and atomic mass units

It is a universal physical constant just as the speed of particle

It is the number of particles that is required to fill a $$1$$-liter container

It is the inverse diameter of an $$H$$ atom

It is the number of electrons in the universe

Solution

Question 3

electrons

protons

neutrons

atoms

none of these

Solution

Question 4

ion

Cation

Anion

Positron

Solution

Question 5

Law of reciprocal proportion

Law of multiple proportion

Law of constant composition

None of these

Solution

Question 6

$$2 : 2 : 3$$

$$1 : 1 : 3$$

$$39 : 36 : 48$$

$$1 : 2 : 6$$

Solution

Question 7

$$0.50$$

$$0.25$$

$$0.75$$

$$0.67$$

Solution

Question 8

1.0 g of butane $$(C_4H_{10})$$

1.0 g of nitrogen $$(N_2)$$

1.0 g of silver ($$Ag$$)

1.0 g of water $$(H_2O)$$

Solution

Question 9

$$9.977\times 10^{-23}gm$$

$$9.8\times 10^{-22}gm$$

$$1\times 10^{-22}gm$$

$$6.6\times 10^{-23}gm$$

Solution

Question 10

$$H_{2}O > H_{2}S > CO_{2} > SO_{2}$$

$$H_{2}O > H_{2}S < CO_{2} > SO_{2}$$

$$H_{2}O < H_{2}S < CO_{2} < SO_{2}$$

$$H_{2}O > H_{2}S > CO_{2} < SO_{2}$$

Solution

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