States Of Matter(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

STATES OF MATTER MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. A glass tube of volume 112 ml containing a gas is partially evacuated till the pressure in it drops to

3.8 \times 10^{- 5}

torr at

0^{0} C

. The number of molecules of the gas remaining in the tube is

3.7×1017
1.5×1014
4.5×1018
1.5×1019

Discuss Question

Answer: Option B. -> 1.5×1014
:
B
PV=nRT

n = \frac{P V}{R T} = \frac{3.8 \times 10^{- 5} \times 112 \times 10^{- 3}}{62.4 \times 273}

= 0.02461 \times 10^{- 8} m o l

No. of molecules =

0.02461 \times 10^{- 8} m o l

= 1.5 \times 10^{14}

Question 22. Which of the given sets of temperature and pressure will cause a gas to exhibit the greatest deviation from ideal gas behavior

100∘ C and 4 atm
100∘ C and 2 atm
-100∘ C and 4 atm
0∘ C and 2 atm

Discuss Question

Answer: Option C. -> -100∘ C and 4 atm
:
C
Gas deviate from ideal gas behaviour to real gas (according to Vander Waal's at low temperature and high pressure)

Question 23. Which of the following exhibits the weakest intermolecular forces

Discuss Question

Answer: Option C. -> He
:
C
Noble gases have the least intermolecular forces (dipole induced dipole)

Question 24. Which of the following has maximum root mean square velocity at the same temperature

Discuss Question

Answer: Option D. -> H2
:
D

V_{r m s} = \sqrt{\frac{3 R T}{M}} ∴ V_{r m s} α \frac{1}{\sqrt{M}}

at same T
because

H_{2}

has least molecular weight so its r.m.s velocity should be maximum.

Question 25. The rate of diffusion of hydrogen and chlorine are in the ratio of 6:1. The density of chlorine as compared to that of hydrogen will be:

Discuss Question

Answer: Option D. -> 36
:
D

\frac{r_{H_{2}}}{r_{C l_{2}}} = \frac{6}{1} = \sqrt{\frac{d_{C l_{2}}}{d_{H_{2}}}}

, If density of hydrogen is 1, then density of chlorine is 36

Question 26. The compressibility factor for nitrogen at 330 K and 800 atm is 1.90 and at 570 K and 200 atm is 1.10. A certain mass of

N_{2}

occupies a volume of 1

d m^{3}

at 330 K and 800 atm. Calculate the volume occupied by same quantity of

N_{2}

gas at 750 K and 200 atm.

Discuss Question

Answer: Option D. -> 4 L
:
D

Z = \frac{P V}{n R T} \Rightarrow 1.90 = \frac{1 \times 800}{n \times R \times 330}

n = \frac{1 \times 800}{1.90 \times R \times 330}

Z = 1.10 = \frac{V \times 200}{n \times R \times 570}

1.10 = \frac{V \times 200 \times 1.90 \times R \times 330}{800 \times R \times 570}

V = 4L

Question 27. Gaseous Benzene reacts with hydrogen gas in presence of a Nickel catalyst to form gaseous cyclohexane according to the Reaction

C_{6} H_{6} + 3 H_{2} ⟶ C_{6} H_{12}

A mixture of

C_{6} H_{6}

and excess

H_{2}

has a pressure of 60mm of Hg in an unknown volume. After the gas had been passed over a Ni catalyst and all the benzene converted to cyclohexane, the pressure of the gas was 30mm of Hg in the same volume at the same temperature. The fraction of

C_{6} H_{6}

(by volume) present in the original volume is

Discuss Question

Answer: Option D. -> 16
:
D
Let initial pressure of

C_{6} H_{6} = P_{1} m m .

And that of

H_{2} = P_{2} m m .

∴ P_{1} + P_{2} = 60 m m . ⟶ (1)

After the Reaction, pressure of

C_{6} H_{6} = 0

(As all has reacted)
Pressure of

H_{2} = P_{2} - 3 P_{1}

Pressure of

C_{6} H_{16} = P_{1}

Total Pressure =

P_{2} - 3 P_{1} + P_{1} = 30 m m

Or,

P_{2} - 2 P_{1} = 30 m m ⟶ (2)

Solving (1) and (2) we get

P_{1} = 10 m m, P_{2} = 50 m m

Fraction of

C_{6} H_{6}

by volume = fraction of moles = fraction of pressure =

\frac{10}{60} = \frac{1}{6}

Question 28. Calculate the pressure of one mole of nitrogen gas in a 10 L cylinder at

50^{\circ} C

using the Van der Waal's equation.Also find the value of ideal gas pressure.(Van der Waal's constants a and b for nitrogen are

1.4 L^{2} a t m m o l^{- 2} a n d 0.04 L^{2} m o l^{- 1}

respectively) .

2.18 atm
2.965 atm
2.652 atm
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
From Van der Waal's equation ,

⟮ p + \frac{a}{V^{2}} ⟯ ⟮ V - b ⟯ = R T

We get ,

⟮ p + \frac{1.4}{100} (10 - 0.04) = 0.0821 \times 323 \Rightarrow p = 2.648 a t m

From ideal gas equation, we have

p = \frac{R T}{V} = \frac{0.0821 \times 323}{10} = 2.652

P_{i d e a l} = 2.648 a t m P_{r e a l} = 2.652 a t m

Question 29. 'x' moles of

N_{2}

gas at S.T.P conditions occupy a volume of 10 litres, then the volume of '2x' moles of

C H_{4}

273^{0} C

and 1.5 atm is

20 lit
26.6 lit
5 lit
16.6 lit

Discuss Question

Answer: Option B. -> 26.6 lit
:
B

\frac{P_{1} V_{1}}{n_{1} T_{1}} = \frac{P_{2} V_{2}}{n_{2} T_{2}}

\frac{1 \times 10}{n \times 273} = \frac{1.5 \times V_{2}}{2 x \times 546}

V_{2} = 26.6 l i t

Question 30. The average speed at

T_{1} K

and most probable speed at

T_{2} K

C O_{2}

gas is

9 \times 10^{4}

cm/sec. The values of

T_{1}

and

T_{2}

are

2143K; 1694K
1726K, 2126K
1684K; 2143K
1684K; 3368K

Discuss Question

Answer: Option C. -> 1684K; 2143K
:
C

¯ C = C_{p}

\sqrt{\frac{8 R T_{1}}{M π}} = \sqrt{\frac{2 R T}{M}}

\frac{8 T_{1}}{π} = T_{2}

\frac{T_{1}}{T_{2}} = \frac{π}{8} = \frac{22}{7} \frac{1}{8} = \frac{22}{56}

T_{1} = 1684 K

T_{2} = 2143 K

← Previous
1
2
3

Latest Videos

Chapter 1 - GLOBAL STEEL SCENARIO & INDIAN STEEL INDUSTRY Part 1 : (13-04-2024) INDUSTRY AND COMPANY AWARENESS (ICA)

Direction Sense Test Part 1 Reasoning (Hindi)

Chapter 1 - RMHP / OHP / OB & BP Part 1 : (14-02-2024) GPOE

Cube & Cuboid Part 1 Reasoning (Hindi)

Data Interpretation (DI) Basic Concept Reasoning (Hindi)

Counting Figures Part 1 Counting Of Straight Lines Reasoning (Hindi)

Sail E0 free webinar Webinars

Real Numbers Part 7 Class 10 Maths

Real Numbers Part 1 Class 10 Maths

Polynomials Part 1 Class 10 Maths

Latest Test Papers

GPOE / GPA Combined 1 Free Mock Test SAIL JO E-0 2024

Chapter 3.1 : INTRODUCTION Chapter 3 : Sinter Plant SAIL E0 - GPOE/GPA 2024

Chapter 2.3 : COAL HANDLING PLANT - Set 1 Chapter 2 : Coke Ovens and Coal Chemicals SAIL E0 - GPOE/GPA 2024

Chapter 1.1 : INTRODUCTION Chapter 1 : Raw Material Handling Plant SAIL E0 - GPOE/GPA 2024

DSP Combined 1 Free Revision Test SAIL JO E-0