Class 11 Exam  >  Class 11 Questions  >  For the following equilibrium PCl5(g) --> PCl... Start Learning for Free
For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm?
Most Upvoted Answer
For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour dens...
Given Data:
- Equilibrium reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
- Vapour density = 100
- 1 mole of PCl5 taken in a 10 dm^3 flask
- Temperature = 300 K

Calculating Moles:
Since the vapour density is given, we can calculate the moles of the gases present in the flask.

- Vapour density (VD) = Molecular weight of the substance / 2
- Molecular weight of PCl5 = 208.25 g/mol
- VD = 100
- Therefore, Molecular weight of the substance = 2 * VD = 2 * 100 = 200 g/mol
- Mass of the substance = Moles of the substance * Molecular weight of the substance
- Mass of PCl5 = 1 mole * 208.25 g/mol = 208.25 g

Calculating Equilibrium Pressure:
The ideal gas equation can be used to calculate the equilibrium pressure.

- PV = nRT
- P = (n/V) * RT
- Since 1 mole of PCl5 is taken in a 10 dm^3 flask, the concentration (n/V) is given by:
- (n/V) = 1 mole / 10 dm^3 = 0.1 mol/dm^3
- R = 0.0821 atm/mol.K (gas constant)
- T = 300 K (temperature)

Substituting the values in the equation:
P = (0.1 mol/dm^3) * (0.0821 atm/mol.K) * (300 K) = 2.463 atm

Answer:
The equilibrium pressure is approximately 2.46 atm, which corresponds to option D.
Community Answer
For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour dens...
Pv=nRT
p×10=1×0.082 ×300
P=2.46
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
Explore Courses for Class 11 exam

Top Courses for Class 11

For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm?
Question Description
For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm?.
Solutions for For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? defined & explained in the simplest way possible. Besides giving the explanation of For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm?, a detailed solution for For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? has been provided alongside types of For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? theory, EduRev gives you an ample number of questions to practice For the following equilibrium PCl5(g) --> PCl3(g) Cl2(g), vapour density is found to be 100 when 1 mole of PCl5 is taken in 10dm3 flask at 300K. Thus equilibrium pressure is A) 2.57 atm B) 1.00atm C) 4.92 atm D) 2.46 atm? tests, examples and also practice Class 11 tests.
Explore Courses for Class 11 exam

Top Courses for Class 11

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev