Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

Consider the chemical equation ,

----- KOH + ----- H₃PO₄ → ----- K₃PO₄ + ----- H₂O.

The equation can be balanced by inserting the following in blank spaces

A
3,1,1,3

B
6.1.2.3

C
3,1,2,3

D
3,2,1,2

Solution

The given equation gets balanced by inserting the coefficients (3 , 1 ,1 ,3 ) in blank spaces from LHS to RHS . Thus, the balanced equation for the given reacion is ,

3KOH + H₃PO₄ → K₃PO₄ + 3H₂O.
Question 2
1 / -0

Consider the chemical reaction gven as,
--CO (g) + -- H₂(g) → -- C₈H₁8(l) + -- H2O.
This equation can be balanced by inserting the following in blank spaces

A
8,2,1,17

B
8,8,1,17

C
8,2,8,17

D
8,17,1,8

Solution

The balanced equation for the given equation is,
8CO(g) + 17H₂(g) ⟶⟶ C₈H₁₈(l) + 8H₂O
hence, the coefficients ( 8,17,1,8) are inserted on blank spaces (from LHS to RHS).
Question 3
1 / -0

Consider the unbalanced chemical equation ,
-- Al + -- O₂ → -- Al₂O₃.
The equation for above reaction can be balanced by inserting the following in blank spaces.

A
4,3,2

B
2,3,1

C
4,2,3

D
2,3,3

Solution

The given equation gets balanced by inserting coefficients (4, 3, 2) in blank spaces from LHS to RHS. The balanced equation is -
4Al + 3O₂ ⟶⟶ 2Al₂O₃
Question 4
1 / -0

^-- Ba₃N₂ +^- H₂O → ^-- Ba (OH)₂ + ^-- NH₃ .

The above equation can be balanced by inserting the following in blank spaces from LHS to RHS respectively.

A
2,6,1,2

B
1,6,3,6

C
1,3,6,2

D
1,6,3,2

Solution

The given equation gets stoichiometrically balanced when the coefficients ( 1, 6, 3, 2 ) starting from LHS to RHS are inserted in given blank spaces.
Thus, the balanced equation for the reaction is,
Ba₃N₂ + 6H₂O ⟶⟶ 3Ba(OH)₂ + 2NH₃
Question 5
1 / -0

The gram molar mass of CO₂ is

A
78 g

B
88 g

C
44 g

D
46 g

Solution

Molar mass of CO₂ = Σ (atomic mass of C , 2*atomic mass of O )

= [12+2(16)] u

Since , gram molar mass

= Molar mass expressed in gms .

∴ gram molar mass of CO₂

= 44 g
Question 6
1 / -0

The molar mass of Al₂O₃ is

A
92 u

B
82 u

C
42 u

D
102 u

Solution

Molar mass

=Σ [ 2 x ( atomic mass of Al ) , 3 x ( atomic mass of O ) ]u

= [(2X27) + (3X16)]u

= ( 54 + 48 ) u

= 102 u.
Question 7
1 / -0

The molar mass of C₆ H₁₀ O₅ is

A
182 u

B
162 u

C
282 u

D
172 u

Solution

Molar massof C₆H₁₀ O₅

= Sum of the atomic masses of six C atoms , ten H atoms & six O atoms

thus , substituting the required atomic masses of

C= 12 ,

H = 1 , &

O = 16 ,

we get ,

Molar mass of C₆ H₁₀ O₅

= [ 6(12)+10(1)+5(16) ] u

= (72 + 10 + 80 ) u

=162 u
Question 8
1 / -0

The gram molar mass of CaCO₃ is

A
1000

B
50 g

C
100 g

D
150 u

Solution

The gram molar mass of Ca CO₃ is calculated by ,

(i) adding up the atomic masses of Ca , C & 3 O atoms   & ,

(ii) representing the molar mass in grams.

Thus , gram molar mass of Ca C O₃

=   Σ [ atomic mass of Ca   , atomic mass of C , 3 x  atomic mass of O    ]

=  [40+12 + (3X16)]g

= ( 40 + 12 + 48 ) g

=100g

It should be noted that ,

atomic mass of Ca

= 12

atomic mass of C

= 12

atomic mass of O

= 16
Question 9
1 / -0

The molar mass of AgNO₃ is

A
169.9 u

B
189.9 amu

C
159.9 u

D
179.9 g

Solution

Molar mass of Ag NO₃

= [ atomic mass of Ag + atomic mass of N + 3 (   atomic mass of O ) ) ] u

Substituting the atomic masses of Ag , N & O as 107.9 , 14 & 16 ,respectively in the above expression we get -

  =  [107.9 + 14 + (3X16)] u

=169.9 u
Question 10
1 / -0

How many atoms of Oxygen are there in 18g of water?

(Hint: Avagadro’s Number

= 6.02 x 10²³ atoms/mol)

A
6.02 x 10²²

B
6.02 x 10²³

C
7.02 x 10²³

D
8.02 x 10²³

Solution

The problem is based upon the concepts of Avogadro's law

.Accordingly , one mole (ie 18 g ) of water contains

= 6.02 x 10²³ molecules of water

. Again , since the number of oxygen atoms in one molecule of water ( H₂ O )

= 1 atom

∴ the number of oxygen atoms in 6.02 x 10²³ molecules of water

.= 6.02 x 10²³ atoms

.

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 8

Result

Some Basic Concepts of Chemistry Test - 8

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SHARING IS CARING

Question 1

3,1,1,3

6.1.2.3

3,1,2,3

3,2,1,2

Solution

Question 2

8,2,1,17

8,8,1,17

8,2,8,17

8,17,1,8

Solution

Question 3

4,3,2

2,3,1

4,2,3

2,3,3

Solution

Question 4

2,6,1,2

1,6,3,6

1,3,6,2

1,6,3,2

Solution

Question 5

78 g

88 g

44 g

46 g

Solution

Question 6

92 u

82 u

42 u

102 u

Solution

Question 7

182 u

162 u

282 u

172 u

Solution

Question 8

1000

50 g

100 g

150 u

Solution

Question 9

169.9 u

189.9 amu

159.9 u

179.9 g

Solution

Question 10

6.02 x 1022

6.02 x 1023

7.02 x 1023

8.02 x 1023

Solution

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6.02 x 10²²

6.02 x 10²³

7.02 x 10²³

8.02 x 10²³