Important Questions for Chemical Science Students - Physical Chemistry

Most Important Questions on Physical Chemistry (Download PDF)

1. Which of the following is true regarding HMO theory?

i) HMO is an absolute method.

ii) HMO treats pi electrons separately from sigma electrons.

iii) Pi electrons determine the properties of the conjugated molecules.

iv) HMO uses only the variation principle.

1. Only iv
2. ii and iv
3. ii and iii
4. only ii

2. The rate constant of a unimolecular reaction was 2.56 x 10-3 and 2.3 x 10-1 at T= 110K and 330K respectively. The rate constant (in s-1 units) at 220K would be:

1. 43 x 10-2
2. 43 x 10-1
3. 4.81 x 10-2
4. 81 x 10-1

3. A system is expanded Reversibly adiabatically from 1 L to 10 L. If initial temp. is 750 K, what would be the final temperature? CP= 29.23 J/K mol.

1. 500 K
2. 200 K
3. 100 K
4. 300 K

4. Which among the following will show maximum flocculation value for Fe(OH)3 Solution?

1. (NH4)3PO4
2. Na2S
3. NH4Cl
4. NaCl

5. Consider the reaction 

Cu2+(aq) + 2e → Cu(s) 

What will be the half-cell potential at 298K where [Cu2+ = 5.0 M and Eº = +0.34V].      

1. 4.6 mV
2. 0.466 V
3. 3.6 V
4. 0.36 V

6. The extent of dissociation of PCl5 at a certain temperature is 20% at one atm pressure. Calculate the pressure at which this substance is half-dissociated at the same temperature.

1. 0.123
2. 0.329

0.420

1. 0.789

7. Improper Axis of symmetry refers to which of the following operations?

1. Rotation
2. Reflection

Rotation followed by perpendicular reflection

1. None of the above

8. Calculate the maximum rotational level which can be occupied at 300K when the rotational constant is 24cm-1.

1. 1
2. 2
3. 4
4. 10

9. The ratio of HCP closed packed atoms to voids in HCP closed packing is:

1. 1:4
2. 2:3
3. 1:6
4. 1:3

10. Two miscible liquids A and B form a solution. Assume that the solution is non-ideal but the vapor above it behaves ideally. For pure A and B, the vapor pressures are 550 torrs and 700 torrs respectively at 35°C. If the total pressure above a solution that is 48-mole percent A, is 500 torr and the mole fraction of A in the vapor is 0.45, determine the activity coefficients of A and B in the solution.

1. 0.854, 0.756
2. 0.765, 0.324
3. 0.854, 0.765
4. 0.845, 0.765

Answer Key:

1. C
2. A
3. D
4. D
5. D
6. A
7. C
8. B
9. D
10. A

Solutions:

Solution 1. HMO calculates the energy and shape of the pi molecular orbitals of the planar conjugated molecules. The calculations are based on the variation principle along with LCAO for pi electrons. It differentiates pi electrons from sigma electrons as pi electrons take part in conjugation. This theory is approximate since it ignores electron-electron repulsions.

Solution 2. By applying the Arrhenius equation:

Solution 3. As the system is reversibly adiabatically expanded, so, 

Given CP = 29.23 J/K mole

Solution 4. Flocculation value or precipitation value is the minimum amount of an electrolyte in millimoles that must be added to one liter of colloidal sols to bring about complete coagulation or precipitation. Coagulation power is inversely proportional to coagulation value, i.e., the smaller the coagulation value of an electrolyte ion, the greater is the coagulating power.

Solution 5. According to Nernst equation:
ECu2+/Cu=EoCu2+/Cu-0.059/n log 1/[Cu2+]

= 0.34 –0.059/2 log (1/5)  

= 0.36 V

Solution 6. Kp = α2P/ (1 – α2)

P = 1 atm, α = 0.2

 Kp = (0.2)2 (1 atm) / (1 – 0.04) = 0.041 atm

Let P' be the pressure at which α = 0.5, then

Kp = α2P'/ (1 – α2)

0.041 atm = (0.5)2P'/ (1 – 0.25)

P = 0.123 atm

Solution 7. A molecule is said to have an improper Axis of rotation of order n if the rotation of 2ᴨ/n about an axis is followed by the reflection in a plane perpendicular to that axis.

Solution 8. The maximum rotational level that can be occupied is:

Here, B = 24 cm-1, T = 300 K, k = 1.38×10-23 J K-1

k (in cm-1) = (1.38×10-23)/(6.626×10-34×3×1010)=0.694 cm-1 K-1

Jmax = 2

Solution 9.  No. of Atoms present in HCP = 6

Total no. voids present in HCP = Td voids + Oh voids

= 12 + 6

= 18

Solution 10. xA, vap = 0.45

Since, the vapor behaves ideally, hence,

pA = xA, vap × P = 0.45 × 500 torr = 225 torr

pB = P – pA = 500 torr – 225 torr = 275 torr

Since the solution behaves non-ideally, Raoult’s law becomes:

aA= 225 torr/550 torr=0.41

Hence,

γA = aA/xA = 0.41/0.48 = 0.854

aB= 275 torr/700 torr=0.393

γB = aB/xB = 0.393/0.52 = 0.756

Download Important Questions On Physical Chemistry- Download PDF Here

Check Out: 

• Study Notes for CSIR NET Life Science - (Download PDF)
• Study Notes for CSIR NET Chemical Science - Download PDF Here!

Hope the above article was helpful for you. Let me know your feedback in the comments section below!!!

More from us:

• Get Unlimited access to Structured Live Courses and Mock Tests - Online Classroom Program
• Get Unlimited access to 60+ Mock Tests - Buy Test Series

BYJU'S Exam Prep Team

Download the BYJU’S Exam Prep App Now. 

The Most Comprehensive Exam Prep App.

#DreamStriveSucceed

App Link: https://bit.ly/3sxBCsm