Some Basic Concepts of Chemistry Test

Result

Some Basic Concepts of Chemistry Test - 36

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

$$3.011\times { 10 }^{ 22 }$$ atoms of an element weights $$1.15 $$ g. The atomic mass of the element is :

A
$$23$$ amu

B
$$10$$ amu

C
$$16$$ amu

D
$$35.5$$ amu

Solution

$$3.011\times { 10 }^{ 22 }$$ atoms of an element weights $$1.15 $$ g.

$$3.011\times { 10 }^{ 22 }=\ 1.15g$$

$$6.023\times { 10 }^{ 23 }=\ x$$?

$$\therefore x\ =\ \dfrac { 1.15 }{ 3.011\times { 10 }^{ 22 } } \times 6.023\times { 10 }^{ 23 }=23$$ amu
Question 2
1 / -0

Mass of beaker=10g
Mass of mixture+beaker=25g
Mixture contains 30% sand. Find the weight of sand.

A
5 g

B
4.5 g

C
3 g

D
3.5 g

Solution

Mass of mixture=$$25g-10g=15g$$
Percentage of sand=30%=$$\frac{Mass of sand}{Mass of mixture}$$
Mass of sand =$$0.3\times15g=4.5 g$$
Question 3
1 / -0

One atomic mass is equal to

A
$$1.66\times10^{-27}$$ g

B
$$1.66\times10^{-24}$$ g

C
$$1.66\times10^{-23}$$ g

D
$$1.66\times10^{-25}$$ g

Solution

One atomic mass unit is equal to $$1.66 \times 10^{-24} g.$$ It is equal to 1/12th of mass of atom of Carbon-12. It is abbreviated as $$\text{u.}$$
Question 4
1 / -0

Which law is not applicable for a nuclear reaction where large amount of energy is released?

A
Law of conservation of mass

B
Law of definite proportion

C
Law of multiple proportion

D
Avagrado's law

Solution

The law implies that during any chemical reaction, nuclear reaction, or radioactive decay in an isolated system, the total mass of the reactants or starting materials must be equal to the mass of the products. Mass conservation remains correct if energy is not lost but if energy is released in a nuclear reaction, the law of conservation of mass does not hold any good.
Question 5
1 / -0

One atomic mass unit is equal to?

(Given, mass of one carbon-12 atom $$= 1.992 \times 10^{-23}$$)

A
$$1.66\times 10^{-27}g$$

B
$$1.66\times 10^{-24}g$$

C
$$1.66\times 10^{-23}g$$

D
$$1.66\times 10^{-25}g$$

Solution

One atomic mass unit (amu or u) is equal to 1/12th of the mass of an atom of carbon-12.

Mass of an atom of carbon-12 $$= 1.992 \times 10^{-23}\ g$$

1 amu $$=$$ 1/12 mass of carbon-12 atom $$= \dfrac {1}{12} \times 1.992 \times 10^{-23}$$ $$= 1.66\times 10^{-24}g$$
Question 6
1 / -0

Avogadro's number is NOT equal to:

A
the number of atoms in 11.2 L of $$\displaystyle { O }_{ 2 }$$ at STP

B
the number of atoms in 1 mole of $$\displaystyle He$$ at STP

C
the number of electrons in 96,500 coulombs

D
the number of $$\displaystyle { SO }_{ 4 }^{ 2- }$$ ions in 1 L of 0.5 M sulphuric acid
Question 7
1 / -0

Which of the following expressions is equal to the number of iron ($$Fe$$) atoms present in $$10.0$$ g $$\displaystyle Fe$$ ? (atomic mass of $$\displaystyle Fe$$ = $$55.9$$ amu)

A
$$ 10\times 55.9\times( 6.022\times { 10 }^{ 23 }) $$ atoms

B
$$\dfrac{( 6.022\times { 10 }^{ 23 })}{10}\times 55.9$$ atoms

C
$$ 10\times \dfrac {( 6.022\times { 10 }^{ 23 } )}{ 55.9}$$ atoms

D
$$\dfrac {55.9}{10} \times ( 6.022\times { 10 }^{ 23 } ) $$

E
$$ \dfrac {10}{ ( 55.9\times 6.022\times { 10 }^{ 23 } )} $$ atoms

Solution

The expression $$\displaystyle 10\times \left( 6.022\times { 10 }^{ 23 } \right)/ 55.9$$ is equal to the number of iron (Fe) atoms present in 10.0 g Fe.
The atomic mass of Fe is 55.9 g/mol.
The mass of Fe is 10.0 g. Mass is divided with atomic mass to obtain number of moles.
The number of moles of Fe $$ = \dfrac {10.0}{55.9}$$ moles.
The number of moles is multiplied with avogadro's number to obtain the number of Fe atoms.
The number of Fe atoms $$ = \dfrac {10.0}{55.9} \times 6.023 \times 10^{23}$$.
Question 8
1 / -0

If $$10$$ liters of CO gas react with sufficient amount of oxygen for a complete reaction, how many liters of $$CO_2$$ gas will be formed?

A
$$5$$

B
$$10$$

C
$$15$$

D
$$20$$

E
$$40$$

Solution

$$\displaystyle 2CO + O_2 \rightarrow 2CO_2 $$

From the balanced equation for the combustion reaction, we can say that the number of moles of $$CO_2$$ formed is equal to the number of moles of $$CO$$ reacted.

Also, volume of a gas is directly proportional to the number of moles.
Hence, the volume of $$CO_2$$ formed is equal to the volume of $$CO$$ reacted.

Hence, if 10 liters of $$CO$$ gas react with sufficient amount of oxygen for a complete reaction, 10 liters of $$CO_2$$ gas will be formed.

Hence, the correct answer is option $$\text{B}$$.
Question 9
1 / -0

How many grams of $$\displaystyle Na$$ are present in $$30$$ g $$\displaystyle NaOH$$?

A
$$10$$ g

B
$$15$$ g

C
$$17$$ g

D
$$20$$ g

E
$$22$$ g

Solution

The molecular weight of $$NaOH$$ is 40 g/mol.
30 g of $$NaOH$$ corresponds to $$\dfrac {30}{40} = 0.75$$ moles.
0.75 moles of $$NaOH$$ contains 0.75 moles of $$Na$$.
The atomic mass of $$Na$$ is 23 g/mol.
0.75 moles of $$Na$$ corresponds to $$23 \times 0.75 = 17$$ g of $$Na$$.
Question 10
1 / -0

A metal nitride contains 28% nitrogen by weight. The molecular formula of metal nitride is $$M_3N_2$$. What is the atomic weight of metal ?

A
72

B
64

C
100

D
24

Solution

100 g of metal nitride will contain 28 g of nitrogen and $$\displaystyle 100-28=72$$ g of metal.
The atomic mass of nitrogen is 14 g/mol.

28 g of nitrogen corresponds to $$\displaystyle \dfrac {28}{14} = 2$$ moles.

The formula of metal nitride is $$\displaystyle M_3N_2$$.

2 moles of N atoms corresponds to 3 moles of metal.

Hence, 72 g of metal is equal to 3 moles.

The atomic weight of metal is $$\displaystyle \dfrac {72}{3}=24 \: g/mol$$.

Hence, option D is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 36

Result

Some Basic Concepts of Chemistry Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

$$23$$ amu

$$10$$ amu

$$16$$ amu

$$35.5$$ amu

Solution

Question 2

5 g

4.5 g

3 g

3.5 g

Solution

Question 3

$$1.66\times10^{-27}$$ g

$$1.66\times10^{-24}$$ g

$$1.66\times10^{-23}$$ g

$$1.66\times10^{-25}$$ g

Solution

Question 4

Law of conservation of mass

Law of definite proportion

Law of multiple proportion

Avagrado's law

Solution

Question 5

$$1.66\times 10^{-27}g$$

$$1.66\times 10^{-24}g$$

$$1.66\times 10^{-23}g$$

$$1.66\times 10^{-25}g$$

Solution

Question 6

the number of atoms in 11.2 L of $$\displaystyle { O }_{ 2 }$$ at STP

the number of atoms in 1 mole of $$\displaystyle He$$ at STP

the number of electrons in 96,500 coulombs

the number of $$\displaystyle { SO }_{ 4 }^{ 2- }$$ ions in 1 L of 0.5 M sulphuric acid

Question 7

$$ 10\times 55.9\times( 6.022\times { 10 }^{ 23 }) $$ atoms

$$\dfrac{( 6.022\times { 10 }^{ 23 })}{10}\times 55.9$$ atoms

$$ 10\times \dfrac {( 6.022\times { 10 }^{ 23 } )}{ 55.9}$$ atoms

$$\dfrac {55.9}{10} \times ( 6.022\times { 10 }^{ 23 } ) $$

$$ \dfrac {10}{ ( 55.9\times 6.022\times { 10 }^{ 23 } )} $$ atoms

Solution

Question 8

$$5$$

$$10$$

$$15$$

$$20$$

$$40$$

Solution

Question 9

$$10$$ g

$$15$$ g

$$17$$ g

$$20$$ g

$$22$$ g

Solution

Question 10

72

64

100

24

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.