JEE Exam  >  JEE Questions  >  At 298 K, the vapour pressure of an ideal sol... Start Learning for Free
At 298 K, the vapour pressure of an ideal solution containing 1 mol of liquid L1 and 2 mol of liquid L2 is 500 mm
Hg. When 2 mol of L1 is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K ?
  • a)
    500 and 500
  • b)
    563 and 469
  • c)
    513 and 494
  • d)
    500 and 1250
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
At 298 K, the vapour pressure of an ideal solution containing 1 molof ...
Case — 1: ideal solution containing one mole of L1 and two moles of L2
Mole fraction of liquid
Mole fraction of liquid
Case — 2: After addition of 2 moles of L1
On solving equation (1) and equation (2) -
Free Test
Community Answer
At 298 K, the vapour pressure of an ideal solution containing 1 molof ...
Given Information:
- Initial vapour pressure of the solution containing 1 mol of L1 and 2 mol of L2 is 500 mmHg.
- When 2 mol of L1 is added to the solution, the vapour pressure increases by 5%.

Solution:

Step 1: Calculate the initial moles of L1 and L2:
- Initial moles of L1 = 1 mol
- Initial moles of L2 = 2 mol

Step 2: Calculate the initial mole fraction of L1 and L2:
- Initial mole fraction of L1 = 1 / (1 + 2) = 1/3
- Initial mole fraction of L2 = 2 / (1 + 2) = 2/3

Step 3: Calculate the initial vapour pressure of L1 and L2:
- Let the vapour pressure of L1 be P1 and L2 be P2.
- Using Raoult's Law, we have:
500 = (1/3) * P1 + (2/3) * P2

Step 4: Calculate the final moles of L1 and L2 after adding 2 mol of L1:
- Final moles of L1 = 1 mol + 2 mol = 3 mol
- Final moles of L2 = 2 mol

Step 5: Calculate the final mole fraction of L1 and L2:
- Final mole fraction of L1 = 3 / (3 + 2) = 3/5
- Final mole fraction of L2 = 2 / (3 + 2) = 2/5

Step 6: Calculate the final vapour pressure of the solution:
- The vapour pressure of the solution increases by 5%, so the new vapour pressure is 500 * (1 + 0.05) = 525 mmHg.
- Using Raoult's Law again, we have:
525 = (3/5) * P1 + (2/5) * P2

Step 7: Solve the two equations to find the respective vapour pressures of L1 and L2:
- Solving the two equations simultaneously, we get:
P1 = 563 mmHg
P2 = 469 mmHg
Therefore, the respective vapour pressures of L1 and L2 in their pure states at 298 K are 563 mmHg and 469 mmHg, respectively. Hence, option (b) is the correct answer.
Explore Courses for JEE exam

Similar JEE Doubts

At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer?
Question Description
At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer?.
Solutions for At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer?, a detailed solution for At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice At 298 K, the vapour pressure of an ideal solution containing 1 molof liquid L1and 2 molof liquid L2is 500 mmHg. When 2 molof L1is added to this solution, the vapour pressure of the solution increases by 5%. What are the respective vapour pressures (in mm Hg) of L1 and L2 in their pure states at 298 K?a)500 and 500b)563 and 469c)513 and 494d)500 and 1250Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

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