Since Avogadro number is $$6.023\times {10}^{23}$$ and atomic mass of nitrogen is $$14$$ amu, one mole of $$N_2$$ gas will have:

$$0.5\times 6.023\times {10}^{23}$$ atoms

$$14\times 6.023\times {10}^{23}$$ atoms

$$28\times 6.023\times {10}^{23}$$ atoms

The smallest particle of matter known

MCQ Questions for Class 9 Chemistry Atoms And Molecules Quiz 5

An ionic compound of element

M

$M$ and chlorine has the formula

M C l_{3}

$MCl_3$ , the molecular mass of

M C l_{3}

$MCl_3$ is 118.5 u. What is the molecular mass of the oxide of an element

M

$M$ ?

0%
28
0%
44
0%
72
0%
99

The difference between ions and atoms is of :

0%
relative size
0%
electronic configuration
0%
presence of charge
0%
All of the above

14 g of element X combine with 16 g of oxygen. On the basis of this information, which of the following is a correct statement?

0%
The element X could have an atomic weight of 7 and its oxide is XO.
0%
The element X could have an atomic weight of 14 and its oxide is $X_2O$ .
0%
The element X could have an atomic weight of 7 and its oxide is $X_2O$
0%
The element X could have an atomic weight of 16 and its oxide is $X_2O$

Explanation

The atomic weight of oxygen is 16. The atomic weight of the element

X

$X$ can be either 7 or 14.

One mole of the oxide will contain one mole (16 g) of oxygen atoms and one mole (14 g, if atomic weight of

X

$X$ is 14) of

X

$X$ atoms. Its formula will be

X O

$XO$ .

One mole of the oxide will contain one mole (16 g) of oxygen atoms and two moles (14 g, if atomic weight of

X

$X$ is 7) of

X

$X$ atoms. Its formula will be

X_{2} O .

$X_2{O}.$

Thus option C is the correct answer.

Smallest particle of an element or a compound which is capable of independent existence is called:

0%
atom
0%
molecule
0%
element
0%
compound

A chemical equation is balanced in accordance with the law of

0%
conservation of mass
0%
multiple proportion
0%
constant proportion
0%
reciprocal proportion

Explanation

There must be the same number of each element on both side of the chemical equation. It adheres with the conservation of mass.

Option

A

$A$ is correct.

Which law stated that 'matter is neither created nor destroyed'?

0%
Law of multiple proportion
0%
Law of conservation of energy
0%
Law of constant composition
0%
Law of conservation of mass

Which of these is not a molecule?

0%
$H_2$
0%
$Cl$
0%
$O_3$
0%
$NaCl$

Explanation

Hint: Molecules are made up of atoms that are held together by chemical bonds.

Option $B$

Explanation of Correct Option:

A molecule is combination of two or more than two atoms.

The symbol

C l

$Cl$ represents an atom of chlorine. The molecule of chlorine is represented by the symbol

C l_{2}

$Cl_2$ .

Explanation of Incorrect Option:

H_{2}

$H_2$ is a symbol of a molecule of hydrogen.

O_{3}

$O_3$ is a symbol of a molecule of ozone.

N a C l

$NaCl$ is a symbol of Sodium Chloride, it is a compound.

The law of conservation of mass was given by :

0%
Dalton
0%
Proust
0%
Lavoisier
0%
Berzelius

One mole of any substance contains

6.023 \times 10^{23}

$6.023\times10^{23}$ particles. It was determined by

0%
Berzilius
0%
Avogadro
0%
Dalton
0%
Perin

Explanation

Avogadro's law states that in one mole of any substance, a total number of

6.023 \times 10^{23}

$6.023 \times 10^{23}$ molecules are present.

So, option

B

$B$ is correct.

Which atom contains exactly 15 neutrons?

0%
$P^{32}$ (atomic number $= 17$ )
0%
$S^{32}$ (atomic number $= 16$ )
0%
$O^{15}$ (atomic number $= 8$ )
0%
$N^{15}$ (atomic number $= 7$ )

Explanation

m a s s n u m b e r = n o . o f p r o t o n s + n o . o f n e u t r o n s

$mass\space number= no. of \space protons + no. of\space neutrons$
Given that mass number of P = 32

Atomic number (no. of electrons

=

$=$ no. of protons)

=

$=$ 17

Number of neutrons

= 32 - 17 = 15

$= 32 - 17 = 15$

Hence, P satisfies the requirement.

Amongst the following, which has the highest molecular mass?

0%
$Cl_2$
0%
$N_2$
0%
$CH_3OH$
0%
$NH_3$

Explanation

The molecular masses of various compounds can be calculated as:

(i)

C l_{2}

$Cl_{2}$ ;

2 \times 35.5

$2\times 35.5$

=

$=$

71 g / m o l

$71 g/mol$

(ii)

N_{2}

$N_{2}$ ;

2 \times 14

$2\times 14$

=

$=$

28 g / m o l

$28 g/mol$

(iii)

C H_{3} O H

$CH_{3}OH$ ;

12 + 3 \times 1 + 16 + 1

$12 + 3\times 1 + 16 + 1$

=

$=$

32 g / m o l

$32 g/mol$

(iv)

N H_{3}

$NH_{3}$ ;

14 + 3 \times 1

$14 + 3\times 1$

=

$=$

17 g / m o l

$17 g/mol$

Thus

C l_{2}

$Cl_{2}$ molecule has the highest molecular mass of

71

$71$ moles.

In a chemical reaction, when 14 g of sodium carbonate reacted with 10g of acetic acid (ethanoic acid), then it was observed that, 16.67 g of sodium acetate solution in water and 7.33 g of carbon dioxide were produced. Which law does this depicts?

0%
Law of definite proportions
0%
Law of conservation of mass
0%
Law of constant proportions
0%
Avogadro's law

Explanation

In the given reaction,

Sodium carbonate (14 g) + Ethanoic acid (10 g) \Rightarrow (S o d i u m e t h a n o a t e + w a t e r) (16.67) + C a r b o n d i o x i d e g a s (7.33 g)

$\underset {(14g)}{\text {Sodium carbonate}}+\underset {(10g)}{\text {Ethanoic acid}}\Rightarrow \underset {(16.67)}{(Sodium ethanoate+water)}+\underset {(7.33 g)}{Carbon dioxide gas}$
Total mass of all reactants combined

= 14 g + 10 g = 24 g

$=14 g+10 g=24 g$
Total mass of all the products formed

= 16.67 g + 7.33 g = 24 g

$= 16.67 g + 7.33 g =24 g$ .
It is observed that Total mass of all reactants combined

= 24 g =

$= 24 g =$ Total mass of all the products formed. Therefore, the given observations are in agreement with the law of conservation of mass.

Which of the following elements has the same molecular mass as its atomic mass?

0%
Nitrogen
0%
Neon
0%
Oxygen
0%
Chlorine

The value of Avogadro number is

0%
$6.25 \times {10}^{18}$
0%
$6.022 \times {10}^{23}$
0%
$96500$
0%
$3.12 \times {10}^{18}$

Explanation

Avogadro number is a constant and it is equals to the number of particles present in 1 mole of a substance and is equals to

6.024 \times 10^{23}

$6.024\times 10^{23}$

Any sample of pure water, irrespective of its source, contains 88.89% oxygen and 11.11 % hydrogen by mass. The data supports the:

0%
Law of conservation of mass
0%
Law of constant composition
0%
Law of multiple proportion
0%
Law of reciprocal proportion

Explanation

The ratio of the mass of hydrogen to the mass of oxygen is

always 1:8, whatever the source of water.

\frac{mass of hydrogen}{mass of oxygen} = \frac{11.11}{88.89} = 1 : 8

$\cfrac {\text { mass of hydrogen}}{\text {mass of oxygen}}=\cfrac {11.11}{88.89}=1:8$

Hence the given data supports law of Constant proportion.

720 g of water contains ______ moles of water.

0%
20
0%
40
0%
60
0%
80

Explanation

Given, mass of

H_{2} O = 720 g

$H_2O = 720\ g$

We know that, molar mass of water =

18 g / m o l

$18 g/mol$

Moles of water

= \frac{Mass of water}{Molar mass of water}

$= \displaystyle \frac{\text{Mass of water}}{\text{Molar mass of water}}$

= \frac{720}{18} = 40 m o l e s

$= \displaystyle \frac{720}{18} = 40\ moles$

Smallest possible unit of a compound which has independent existence is :

0%
molecule
0%
atom
0%
ion
0%
electron

The mass of

N a

$Na$ present in 11.7 g of

N a C l

$NaCl$ is:

0%
2.3 g
0%
4.6 g
0%
6.9 g
0%
7.1 g

Explanation

Molar mass of NaCl is 58 g

58 g of NaCl has

N a = 23 g

$Na=23 \ g$

11.7g of NaCl has Na=

\frac{23}{58} \times 11.7 = 4.63 g

$\dfrac{23}{58}\times{11.7}=4.63g$

Option B is correct.

Since Avogadro number is

6.023 \times 10^{23}

$6.023\times {10}^{23}$ and atomic mass of nitrogen is

14

$14$ amu, one mole of

N_{2}

$N_2$ gas will have:

0%
$0.5\times 6.023\times {10}^{23}$ atoms
0%
$2\times 6.023\times {10}^{23}$ atoms
0%
$14\times 6.023\times {10}^{23}$ atoms
0%
$28\times 6.023\times {10}^{23}$ atoms

Explanation

1 molecule of

N_{2}

$N_2$ = 2 atoms of

N

$N$

1 mole of

N_{2} = 2 \times N_{A}

$N_2 = 2 \times N_A$ atoms of

N

$N$ (since, one mole

= 6.023 \times 10^{23}

$= 6.023 \times 10^{23}$ number of entities)
1 mole of

N_{2}

$N_2$ gas

= 2 \times 6.023 \times 10^{23}

$= 2\times 6.023 \times 10^{23}$ atoms of

N

$N$ .

Radius of atom is of the order :

0%
$10^{-10}-10^{-8}cm$
0%
$10^{-12}-10^{-14}cm$
0%
$10^{-14}-10^{-16}cm$
0%
$10^{-16}-10^{-20}cm$

Explanation

Hint: It is based on experimental data observations of Rutherford's scattering experiment.

Correct Answer: Option

A

$A$

Explanation of Correct Option:

On the basis of scattering experiments, Rutherford proposed the model of an atom, which is known as the nuclear atomic model.

According to this model -

(i) An atom consists of a heavy positively charged nucleus where all the protons are present.

(ii) The volume of the nucleus is very small and is only a minute fraction of the total volume of the atom.

Nucleus has a radius of the order of

10^{- 13}

$10^{-13}$ cm and the atom has a radius of the order of

10^{- 8}

$10^{-8}$ cm.

Explanation of Incorrect Option:

Other options fall out of range of radius so they are incorrect.

One atomic mass unit is equivalent to

0%
$1.67\times 10^{-27} g$
0%
$1.66\times 10^{-27} kg$
0%
$1.67\times 10^{-27} mg$
0%
$1.6\times 10^{-24} g$

Explanation

One amu

= \frac{1}{N_{A}} = \frac{1}{6.02 \times 10^{23}}

$=\frac {1}{N_A}=\frac {1}{6.02\times 10^{23}}$

= 1.66 \times 10^{- 24} g = 1.66 \times 10^{- 27} k g

$=1.66\times 10^{-24}g=1.66\times 10^{-27}kg$

So, the correct options are

B

$B$ and

D

$D$

2.40 g of the element

Z

$Z$ combines exactly with 1.6 g of oxygen to form a compound with the formula

Z O_{2}

$ZO_2$ . What is the relative atomic mass of

Z

$Z$ ?

0%
24.0
0%
32.0
0%
48.0
0%
64.0

The mass of one atom of carbon-12 is:

0%
$1\ g$
0%
$12\ g$
0%
$12\times 6.02\times 10^{23}\ g$
0%
$\dfrac {12}{6.02\times 10^{23}}\ g$

Explanation

One mole of any element will be equal to its molar mass or atomic weight in grams.

Atomic mass of carbon = 12 g

So, one mole of carbon atom will weigh 12 g

1 mole of atoms =

6.02 \times 10^{23}

$6.02 \times 10^{23}$ atoms

So, mass of Carbon−12 = 12 g =

6.02 \times 10^{23}

$6.02 \times 10^{23}$ atoms

Mass of 1 atom =

\frac{12}{6.02 \times 10^{23}}

$\dfrac {12}{6.02 \times 10^{23}}$ g

An atom is:

0%
The smallest particle of matter known
0%
The smallest particle of a gas
0%
The smallest indivisible particle of an element that can take part in a chemical change
0%
Radioactive emission

Explanation

An atom is the smallest indivisible particle of an element that can take part in a chemical change.

Greek word atom means indivisible. It is the ultimate particle of matter.

So, the correct option is

C

$C$

Atomic theory was given by:

0%
John Dalton
0%
Neils Bohr
0%
E. Rutherford
0%
Haber Bosch

Explanation

(A)

$(A)$ John Dalton

R e a s o n

$Reason$ : Atomic theory was given by John Dalton, represented as,

1

$1$ Everything is composed of atoms, which are the building blocks of matter and cannot be destroyed.

2

$2$ All atoms of an element are identical.

3

$3$ The atoms of different elements vary in size and mass.

4

$4$ Compounds are produced through different whole number combinations of atoms.

5

$5$ A chemical reaction results in rearrangement of atoms in the reactant and product compounds.

'All matter is composed of very small particles called Parmanu' was first suggested by:

0%
John Dalton
0%
J.J. Thomson
0%
Kanad
0%
William J. Crooke

Which of the following contains the largest number of molecules?

0%
0.2 mole of H $_2$
0%
8.0 mole of H $_2$
0%
17 g of H $_2$ O
0%
6.0 g of CO $_2$

Explanation

1 mole of any substance contains

6.02 \times 10^{23}

$6.02 \times 10^{23 }$ (Avogadro's number ) of molecules.

A. 0.2 mole of H

_{2}

$_2$ contains

0.2 \times 6.02 \times 10^{23} = 1.2 \times 10^{23}

$0.2 \times 6.02 \times 10^{23 }=1.2 \times 10^{23 }$ molecules.

B. 8.0 mole of H

_{2}

$_2$ contains

8.0 \times 6.02 \times 10^{23} = 4.8 \times 10^{24}

$8.0 \times 6.02 \times 10^{23 }=4.8 \times 10^{24 }$ molecules.

C. 17 g of H

_{2}

$_2$ O

= \frac{17 g}{18 g/mol} = 0.94 moles

$\displaystyle =\dfrac { \text {17 g } }{\text { 18 g/mol}} = \text { 0.94 moles}$

17 g of H

_{2}

$_2$ O contains

0.94 \times 6.02 \times 10^{23} = 5.7 \times 10^{23}

$0.94 \times 6.02 \times 10^{23 }=5.7 \times 10^{23 }$ molecules.

D. 6.0 g of CO

_{2}

$_2$

= \frac{6.0 g}{44 g/mol} = 0.136 moles

$\displaystyle =\dfrac { \text {6.0 g } }{\text { 44 g/mol}} = \text { 0.136 moles}$

6.0 g of CO

_{2}

$_2$ contains

0.136 \times 6.02 \times 10^{23} = 8.2 \times 10^{22}

$0.136 \times 6.02 \times 10^{23 }=8.2 \times 10^{22 }$ molecules.

Hence, 8.0 mole of H

_{2}

$_2$ contains the largest number of molecules

Which of the following weighs the most?

0%
$10^{23}$ molecules of $H_2$
0%
1 mole of $H_2$ O
0%
1 mole of $N_2$
0%
$10^{22}$ molecules of $O_2$

Explanation

Molecular weight of

H_{2} = 2

$H_2=2$ g/mol. 1 mole of

H_{2}

$H_2$ weighs 2 g and contains

6.02 \times 10^{23}

$6.02 \times 10^{23 }$ (Avogadro's number ) molecules.

Weight of

10^{23}

$10^{23}$ molecules of H

_{2}

$_2$

= \frac{1 \times 10^{23}}{6.02 \times 10^{23}} \times 2 = 0.33

$\displaystyle = \dfrac {1 \times 10^{23 }}{6.02 \times 10^{23 }} \times 2=0.33$ g

Weight of 1 mole of H

_{2}

$_2$ O

=

$=$ molecular weight

= 2 (1) + 16 = 18

$\displaystyle =2(1)+16 =18$ g.

Weight of 1 mole of N

_{2}

$_2$

=

$=$ molecular weight

= 2 (14) = 28

$\displaystyle =2(14)=28$ g.

Molecular weight of

O_{2} = 32

$O_2=32$ g/mol. 1 mole of

O_{2}

$O_2$ weighs 32 g and contains

6.02 \times 10^{23}

$6.02 \times 10^{23 }$ (Avogadro's number ) molecules.

Weight of

10^{22}

$10^{22}$ atoms of oxygen

= \frac{1 \times 10^{22}}{6.02 \times 10^{23}} \times 32 = 0.53

$\displaystyle = \dfrac {1 \times 10^{22 }}{6.02 \times 10^{23 }} \times 32=0.53$ g

Hence, 1 mole of N

_{2}

$_2$ weighs the most.

6.4 g of oxygen will contain ______ number of oxygen molecules.

0%
$6.023 \times 10^{23}$
0%
$6.023 \times 10^{22}$
0%
$1.2046 \times 10^{23}$
0%
$1.2046 \times 10^{22}$

Explanation

Molar mass of oxygen

= 32 g / m o l

$= 32 g/mol$

Moles in 6.4 g oxygen

= \frac{6.4}{32} = 0.2

$= \dfrac{6.4}{32} = 0.2$ moles
Number of

O_{2}

$O_2$ molecules in 0.2 mole

= 0.2 \times 6.023 \times 10^{23}

$\displaystyle = 0.2\times 6.023\times 10^{23}$

= 1.2046 \times 10^{23}

$=1.2046\times 10^{23}$ molecules

Which of the following statement is wrong?

0%
An atom is electrically neutral.
0%
The size of a cation is smaller than that of the corresponding atom.
0%
The size of an anion is bigger than that of the corresponding atom.
0%
An atom and its ion have an unequal number of protons.

Which of the following compound has formula mass of 322 u?

(H = 1, C = 12, 0 = 16, N a = 23, S = 32, C u = 63.5)

$(H= 1, C=12, 0=16, Na=23, S=32, Cu=63.5)$

0%
$Na_2CO_3.10H_2O$
0%
$CuSO_4.5H_2O$
0%
$Na_2CO_3.9 H_2O$
0%
$Na_2SO_4.10 H_2O$

Explanation

${ Na }_{ 2 }{ CO }_{ 3 }.10{ H }_{ 2 }O=286u\\ { CuSO }_{ 4 }.{ 5H }_{ 2 }O=249.5u\\ { Na }_{ 2 }{ CO }_{ 3 }.{ 9H }_{ 2 }O=268u\\ { Na }_{ 2 }{ SO }_{ 4 }.10{ H }_{ 2 }O=322u$

Hydrogen combines with nitrogen in a 3:14 weight ratio to form ammonia. If every molecule of ammonia contains three atoms of hydrogen and one atom of nitrogen, an atom of nitrogen must weigh :

0%
14 times the mass of a hydrogen atom
0%
14/3 times the mass of a hydrogen atom
0%
3 times the mass of a hydrogen atom
0%
3/14 times the mass of a hydrogen atom

Explanation

In ammonia, nitrogen and hydrogen are always present in the ratio 14:3 by mass

Let's assume mass of a Nitrogen atom is

$X$ and that of a Hydrogen atom is

$Y$ .

From the law of constant proportion,

$3Y:X = 3:14 \Rightarrow X=14 \times Y$

Mass of a Nitrogen atom

$(X)$ is 14 times that of a Hydrogen atom (

$Y$ ).

Hence Option A is correct.

The sum of masses of the atoms in the molecules is known as :

0%
Atomic mass
0%
Molecular mass
0%
Gram molecular weight
0%
None of the above

The weight of one molecule of a compound of molecular formula

$C_{60} H_{122}$ is:

0%
$1.2\times 10^{-20} g$
0%
$5.025\times 10^{-23} g$
0%
$1.4\times 10^{-21} g$
0%
$6.023\times 10^{-20} g$

Explanation

Weight of 1 mole of

$C_{60}H_{122} = M.W.= 60\times 12 + 122 \times 1=842$

Number of molecules in 1 mole

$C_{60}H_{122} = 6.023 \times 10^{23}$

Weight of one molecule

$= \dfrac {842}{6.02\times 10^{23}} gm$

$= 140\times 10^{-23} g = 1.4\times 10^{-21} g$ .

Hence, the correct option is

$\text{C}$

Formula unit mass of

$NaHCO_{3}$ is :

0%
$84\ u$
0%
$76\ u$
0%
$52\ u$
0%
$64\ u$

Explanation

Formula unit mass of

$NaHCO_{3} =$ Mass of

$Na +$ Mass of

$H+$ Mass of

$C+$ Mass of

$O$

$= 23\ u + 1\ u + 12\ u + 48\ u$

$= 84\ u$

Which among the following has molecular mass of 36.5 u?

0%
Carbon molecules
0%
Hydrogen molecules
0%
Sulphur molecules
0%
Hydrochloric acid

Explanation

Molar mass of Hydrochloric acid

$=1\times$ atomic mass of

$H + 1\times$ atomic mass of

$Cl$

$= 1\times (1) + 1\times (35.5)$

$=36.5\ u$

Molecular mass of nitric acid is:

0%
$22\ u$
0%
$12\ u$
0%
$33\ u$
0%
$63\ u$

Explanation

Molecular mass of nitric acid

$(HNO_3)$

$= (1 \times atomic\ mass\ of\ H) + (1 \times atomic\ mass\ of\ N) + (3 \times atomic\ mass\ of\ O)$

$= 1\times (1\ u) + 1\times (14\ u) + 3\times (16\ u)$

$= 63\ u$

Hydrogen combines with oxygen in a 1:8 weight ratio to form water. If every molecule of water contains two atoms of hydrogen and one atom of oxygen, an atom of oxygen must weigh:

0%
8 times the mass of a hydrogen atom
0%
16 times the mass of a hydrogen atom
0%
1/16 times the mass of a hydrogen atom
0%
1/8 times the mass of a hydrogen atom

Explanation

The ratio of the mass of hydrogen to the mass of oxygen is always 1:8.

Let's assume the mass of an Oxygen atom is

$X$ u and that of a Hydrogen atom is

$Y$ u.

From the law of constant proportion,

$2Y:X = 1:8 \Rightarrow X=16 \times Y$

The mass of an Oxygen atom

$(X)$ is 16 times that of a Hydrogen atom (

$Y$ ).

Hence Option B is correct.

1.008 grams of hydrogen combines with 35.453 g of chlorine to form 36.463 g of a pure compound (hydrogen chloride). Dalton's explanation for this experimental fact might be:

0%
One atom of hydrogen combines with 35.453 atoms of chlorine in this reaction
0%
Hydrogen and chlorine atoms always combine in a 1:35 ratio.
0%
Chlorine accepts hydrogen's electron to form polar covalent hydrogen chloride
0%
Hydrogen and chlorine atoms aren't created or destroyed in the process so the reactant mass is the same as the product mass.

The symbol of copper is:

0%
$Co$
0%
$Cr$
0%
$Cp$
0%
$Cu$

What is the atomic mass (u) of chlorine?

0%
34
0%
35.5
0%
33
0%
35

The illustrations show a conservation-of-mass experiment. The solution in the beaker lost mass because

0%
materials have less mass at high temperatures
0%
the mass of the reactants and products was less than $100\ g$
0%
sodium sulfate $(Na_{2}SO_{4})$ is lighter than air
0%
some of the water molecules turned into gas

In the procedure shown above, a calcium chloride solution is mixed with a sodium sulfate solution to create the products shown. Which of the following is illustrated by this activity?

0%
Law of conservation of mass
0%
Law of definite proportions
0%
Dalton's law atomic theory
0%
Theory of covalent bonding

Mass can neither be created nor destroyed. This is in agreement with which law of chemical combination?

0%
Law of definite proportion
0%
Law of multiple proportion
0%
Law of conservation of mass
0%
Law of conservation of momentum

Compounds formed by metals and non-metals contain charged species which are known as:

0%
cations
0%
anions
0%
Both A and C
0%
None of the above

Explanation

Compounds formed by metals and non-metals contain charged species which are known as ions. Such compounds are also known as ionic compounds. For example,

$NaCl$ - it contains sodium ion (cation) and chloride ion (anion).

So, option C is correct.

Which is not one of the laws of chemical combinations?

0%
Law of multiple proportion
0%
Law of conservation of mass
0%
Law of conservation of energy
0%
Law of definite proportion

Calculate the formula unit mass (u) of

$\displaystyle { Na }_{ 2 }O$ .

$Atomic \ mass \ of \ Na = 23 \ u, O=16 \ u$

0%
$60$
0%
$62$
0%
$61$
0%
$59$

Explanation

Formula unit mass (u) of

$\displaystyle { Na }_{ 2 }O$ = 2 x atomic mass of sodium + atomic mass of oxygen

$= 2 \times 23 + 16 = 62 \ u$

In a chemical reaction,

$100 g$ baking soda mixture containing sodium bicarbonate and vinegar on heating gives

$43 g$ of carbon dioxide gas. What mass of solid residue is left in food?

0%
$54\ g$
0%
$55\ g$
0%
$56\ g$
0%
$57\ g$

Explanation

According to law of conservation of mass, total mass of reactants is equal to total mass of products. Here baking soda mixture on heating gives solid residue and carbon dioxide.

$M_{\text {Baking soda mixture}} = M_{\text {solid residue}} + M_{CO_{2}}$

or it can be written as:

$M_{\text {solid residue}} = M_{\text {Baking soda mixture}} - M_{CO_{2}}$

$(Eq 1)$

According to the question

$M_{\text {Baking soda mixture}} = 100 g$

$M_{CO_{2}} = 43 g$

Substituting the value in equation

$1$ , we get

The mass of solid residue is

$100g - 43g =57 g$

The formula unit mass of

$X$ is

$342\ u .$ Among the following, X is :

0%
$H_2SO_4$
0%
$PbI_2$
0%
$Al_{2} (SO_{4})_{3}$
0%
$Al(OH)_3$

Explanation

Formula unit mass of

$Al_{2}(SO_{4})_{3} = (2\times\ {Atomic \ mass\ of\ Al}) + (\text 3\times{Mass\ of\ S}) + (12\times\ {Mass\ of\ O}$ )

$= 2\times 27\ u + 3\times 32\ u + 12\times 16\ u$

$=342\ u$

Hence,option

$C$ is correct.

Formula unit mass of

$CaCO_{3}$ is:

0%
$104\ u$
0%
$106\ u$
0%
$100\ u$
0%
$1144\ u$

Explanation

Formula unit mass of

$CaCO_{3} = \text {Atomic mass of Ca} + \text {Atomic mass of C} + (\text {Atomic mass of O} \times 3)$

$= 40\ u + 12\ u + (16\ u\times3)$

$= 100\ u$

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 5 - MCQExams.com

0:0:3

Practice Class 9 Chemistry Quiz Questions and Answers

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 5 - MCQExams.com

0:0:3

Practice Class 9 Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.