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

12th Grade > Chemistry

MOLE CONCEPT MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. Calculate the volume of

C O_{2}

evolved at NTP when 20g of

C a C O_{3}

is heated strongly.

3.38 L
4.48
5.58 L
6.68 L

Discuss Question

Answer: Option B. -> 4.48
:
B
The question gives weight of reactants and asks the volume of product. So, it is a weight-volume problem. Start by balancing the reaction

C a C O_{3}

\to

CaO +

C O_{2 (g)}

↑

∴

1 mole of

C a C O_{3}

gives 1 mole of

C O_{2}

From Avagadro's Law, 1 mole of

C O_{2}

at NTP occupies 22.4L

∴

1 mole of

C a C O_{3}

= 22.4 L of

C O_{2}

100g = 22.4L of

C O_{2}

∴

20g = 20

\times

\frac{22.4}{100}

= 4.48 L of

C O_{2}

Question 12. The mass of an atom of C is

1 g
1.99×10−23g
112
1.99×10−20g

Discuss Question

Answer: Option B. -> 1.99×10−23g
:
B

6.022 \times 10^{23}

C atoms = 12 g
1 C atom =

\frac{12}{6.022 \times 10^{23}} = 1.99 \times 10^{- 23} g

Question 13. An analytic balance has uncertainty In measurement equal to

\pm

1 mg. Then report the result in terms of percentage if the weight of a compound is
(a) 1 g
(b) 10 g
(c) 100 g

1±0.1%;10±0.01%;100 ±0.001%
1±0.2%;10 ±0.02%;100 ±0.002%
1±0.1%;10 ±0.1%;100 ±0.1%
1±0.2%;10 ±0.2%;100 ±0.002%

Discuss Question

Answer: Option A. -> 1±0.1%;10±0.01%;100 ±0.001%
:
A
(a)

1 \pm 0.1 %

(b)

10 \pm 0.01 %

(c)

100 \pm 0.001 %

The uncertainty in measurement is expressed in terms of percentage by putting

\pm

sign before it, e.g.,

150 \pm 1 %

, et. Smaller the quantity to be measured, greater is the percentage uncertainty, and the instrument should be more precise for the measurement of smaller quantities.

1 m g = \frac{1}{1000} g \frac{1}{1000} \times 100 = \frac{1}{10} = 0.1 1 \pm 0.1 %

Question 14. Which among the following has the highest normality?

8 g of KOH per 100 ml of solution
0.5 M sulphuric acid
1 N phosphoric acid
2 moles of NaOH per 1000 ml of solution

Discuss Question

Answer: Option D. -> 2 moles of NaOH per 1000 ml of solution
:
D

N = \frac{no.of equivalents}{Volume of solution(L)}

N of KOH =

\frac{[\frac{8}{56}]}{0.1} = 1.43 N

N of

H_{2} S O_{4}

= M

\times

n factor =

0.5 \times 2 = 1 N

equivalents of NaOH = moles

\times

n factor
=

2 \times 1 = 2

eq
N of NaOH =

\frac{2}{1} = 2

Question 15. How many years it would take to spend Avogadro's number of rupees at the rate of 1 million rupees in one second?

19.098×1019 years
19.098 years
19.098×109 years
None of these

Discuss Question

Answer: Option C. -> 19.098×109 years
:
C

10^{6}

rupees are spent in 1sec.

∴ 6.022 \times 10^{23}

rupees are spent in

= \frac{(1 \times 6.022 \times 10^{23})}{(10^{6} \times 60 \times 60 \times 24 \times 365)} y e a r s

= 19.098 \times 10^{9} y e a r

Question 16. What volume of 0.10 M

H_{2} S O_{4}

must be added to 50 ml of a 0.10 M NaOH solution to make a solution in which the molarity of the

H_{2} S O_{4}

is 0.050 M?

400 ml
50 ml
100 ml
150 ml

Discuss Question

Answer: Option B. -> 50 ml
:
B
2NaOH +

H_{2} S O_{4}

⟶

N a_{2} S O_{4}

+ 2

H_{2} O

0.0025 moles of

H_{2} S O_{4}

will neutralise 0.005 moles NaOH.
So for 0.01 V moles H2SO4 there will be 0.01v-0.0025 moles H2SO4 left unused in the total volume(v+50ml) of solution .
The concentration here will be

\frac{0.01 - 0.0025}{v + 50}

\times

1000 = 0.05M

\frac{100 V - 2.5}{V + 50}

=0.05

\frac{100 V - 2.5}{V + 50}

= 2.5+2.5
99.95V= 5
V= 0.05002L
V= 50.02ML

Question 17. Avogadro’s number represents the number of atoms in

12g of C -12
32g of sulphur
32g of oxygen
12.7g of iodine

Discuss Question

Answer: Option A. -> 12g of C -12
:
A
It is the total number of particles (atoms, molecules) present in

C_{12}

isotope

Question 18.

\begin{matrix} I s o t o p e & R e l a t i v e A b u d a n c e (%) & A t o m i c M a s s (a m u)^{12} C & 98.892 & 12^{13} C & 1.108 & 13.00335^{14} C & 2 \times 10^{- 10} & 14.00317 \end{matrix}

From the data given in the table, calculate the average atomic mass of C

13 u
12.011 u
12.001 u
12.0011 u

Discuss Question

Answer: Option B. -> 12.011 u
:
B
Average atomic mass is the weighted average of atomic mass of all the isotopes.
Contribution by each isotope = (Relative abundance) (Atomic mass)
Contribution by

_{6}^{12} C

= (0.98892)(12)/100 = 11.86704

_{6}^{13} C

= (0.01108)(13.00335)/100 = 0.14407712

_{6}^{14} C

= (2

\times

10^{- 10}

)(14.003170/100 = 28.00634

\times

10^{- 12}

∴

Average atomic mass = sum of all contribution = 12.011 u

Question 19. 10g of Fe powder was mixed with a

C u S O_{4}

solution which contained 70g of

C u S O_{4}

. What is the weight of the copper obtained?
[Fe = 55.85g Cu = 63.6g]

10.39
11.39
12.39
13.39

Discuss Question

Answer: Option B. -> 11.39
:
B
The question talks about reaction between Fe and

C u S O_{4}

. Observe that weights of the reactants are given and the weight of a product is asked. Thus, it is a weight-weight stoichiometry problem.
In all such stoichiometry problems, start by balancing the reaction
Fe +