Mole Concept(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

MOLE CONCEPT MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. Calculate n-factor of

N a_{3} P O_{4}

in the following reaction.

N a_{3} P O_{4}

3 B a C l_{2}

\to

6NaCl +

B a_{3} (P O_{4})_{2}

Discuss Question

Answer: Option B. -> 3
:
B
This is the type of reaction where none of the atoms involved undergo a change in their oxidation states.
A simple way to find the n-factor of

N a_{3} P O_{4}

in such reactions would be to find the amount of cationic charge replaced per molecule.
In

2 N a_{3} P O_{4}

, we have

{6 N a}^{+}

So,

N a_{3} P O_{4}

has

{3 N a}^{+}

{N a}^{+}

can be cosider to be similar to

H^{+}

H_{3} P O_{4}

(just an analogy)

Question 22. How many grams of oxygen are required for complete combustion of 2 moles of benzene?

450g
480g
500g
530g

Discuss Question

Answer: Option B. -> 480g
:
B
The balanced reaction for combustion of benzene is

2 C_{6} H_{6} + 15 O_{2} \to

12 C O_{2} + 6 H_{2} O

Therefore, 15 moles of oxygen gas is required for complete combustion of benzene.
We know that,

1 m o l e

O_{2}

32 g

∴

15 m o l e s

O_{2}

15 \times 32

480 g

Question 23. Maximum number of atoms are present in

11.2 litresof SO2at STP
22.4 litres of Helium at STP
2.0 g. of hydrogen
11.2 litres of methane at STP

Discuss Question

Answer: Option D. -> 11.2 litres of methane at STP
:
D
No. of atoms in

11.2 l i t .

S O_{2}

\frac{11.2}{22.4} \times N_{A} \times 3 = 1.5 N

No. of Atoms in

2 g

H_{2}

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

No. of Atoms in

22.4 l i t .

H e

\frac{22.4}{22.4} \times N_{A} = N

No. of Atoms in

11.2 l i t

C H_{4}

\frac{11.2}{22.4} \times N_{A} \times 5

2.5 N

Question 24. The mass of CaO that would be obtained by heating 20 kg of

90 %

pure lime stone is:

11.2kg
8.4kg
10.08kg
16.8kg

Discuss Question

Answer: Option C. -> 10.08kg
:
C
Mass of pure limestone =

(\frac{90}{100}) \times 20 = 18 k g

C a C O_{3} Δ - \to C a O + C O_{2}

100 kg of

C a C O_{3}

forms 56 kg of CaO
18 kg of

C a C O_{3}

forms

\frac{(18 \times 56)}{100} = 10.08 kg ofCaO

Question 25. Calculate the normality of 1.11 gm of

C a C l_{2}

is dissolved in 250 ml solution.

0.1 N
0.25 N
0.08 N
0.3 N

Discuss Question

Answer: Option C. -> 0.08 N
:
C
Normality =

\frac{N o . o f g r a m e q u i v a l e n t s o f s o l u t e}{V o l u m e o f s o l u t i o n (i n l i t r e)}

equivalent mass =

\frac{M o l e c u l a r M a s s}{V a l e n e c y f a c t o r}

\frac{111}{2}

[

{C a}^{+ 0}

\to

{C a}^{2 +}

2 e -

] = 55.5
No of gram equivalents =

\frac{1.11}{55.5}

\Rightarrow

Normality =

\frac{\frac{1.11}{55.5}}{\frac{250}{1000}}

= 0.08 N

Question 26. What is the molecular mass of a substance, each molecule of which contains 9 carbon atoms, 13 hydrogen atoms and

2.33 \times 10^{- 23}

g of another component?

135.04
100
111
28

Discuss Question

Answer: Option A. -> 135.04
:
A
The molecule has C, H and other component. Mass of 9 C atoms =

12 \times 9 = 108

amu
Mass of 13 H atoms =

13 \times 1 = 13

amu value of 1 amu
Mass of other component =

\frac{(2.33 \times 10^{- 23})}{1.66 \times 10^{- 24}}

∴

Total mass of one molecule =

108 + 13 + 14.4 = 135.04

amu

∴

Mol. Mass of substance = 135.04

Question 27. Ordinary water contains one part of heavy water per 6000 parts by weight.The number of heavy water molecules present in a drop of water of volume 0.01 ml is (Density of water is 1 g/ml)

2.5 × 1016
2.5 × 1017
5 × 1016
7.5 × 1016

Discuss Question

Answer: Option C. -> 5 × 1016
:
C

\frac{H_{2} O}{6000}

\frac{D_{2} O}{1}

0.01 x

x = \frac{1 \times 0.01}{6000} = 1.66 \times 10^{- 6}

20 g m \to 6.023 \times 10^{23}

1.66 \times 10^{- 6} \to x

x = \frac{1.66 \times 10^{- 6} \times 6.023 \times 10^{23}}{20} = 5 \times 10^{16}

Question 28. Express 6^' 10” in SI units.

2.06 m
2.07 m
2.09 m
2.08 m

Discuss Question

Answer: Option D. -> 2.08 m
:
D
1 Feet = 12 inches

∴

6^' 10^” = (12

\times

6) + 10 = 82 inches
1 inch = 2.54 cm

∴

82 inches = (82

\times

2.54) cm
(centi =

10^{- 2}

) = 208.28 cm = 208.28

\times

10^{- 2}

m
= 2.0828 m

∴

6^' 10^” = 2.08 m

Question 29.

2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}

If we start with 2g of VO and 5.75g of

F e_{2} O_{3}

, which is the limiting reagent?
[V = 47.87g] [Fe = 55.85g]

VO
Fe2O3
Fe2O
V2O5

Discuss Question

Answer: Option B. -> Fe2O3
:
B
If you check, the reaction is already balanced 2 moles of VO = 3 moles of

F e_{2} O_{3}

2 \times (47.87 + 16) = 3 \times (2 \times 55.85 + 3 \times 16)

2 (63.87) g = 3 \times (159.7) g

127.74 g

V O

478.1 g

F e_{2} O_{3}

∴

2 g

V O

\frac{2 \times 478.1}{127.74} = 7.50 g

F e_{2} O_{3}

But only

5.75 g

F e_{2} O_{3}

is present.
This

F e_{2} O_{3}

will get used up in the reaction and is the limiting reagent.

Question 30. Acetylene is the first member of a family of hydrocarbons called alkynes.

H - C \equiv C - H

What is its empirical formula?

CH
C2H2
(CH)2
(CH)3

Discuss Question

Answer: Option A. -> CH
:
A
Empirical formula is made up of simplest whole number ratio of all the elements of the molecule.
In the given case, acetylene is