Chemistry Exam  >  Chemistry Questions  >  A mixture of nitrogen and water vapours is ad... Start Learning for Free
A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:
  • a)
    443.34 L
  • b)
    246.3 L
  • c)
    12.315 L
  • d)
    24.63 L
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A mixture of nitrogen and water vapours is admitted to a flask at 760 ...
Ans.

View all questions of this test
Most Upvoted Answer
A mixture of nitrogen and water vapours is admitted to a flask at 760 ...
To solve this problem, we need to use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for V:

V = (nRT) / P

First, we need to calculate the number of moles of nitrogen and water vapors present in the flask. To do this, we can use the partial pressure of nitrogen and the total pressure of the mixture.

Partial pressure of nitrogen (P_nitrogen) = Total pressure - Partial pressure of water vapors

P_nitrogen = 760 torr - 722 torr = 38 torr

Next, we need to convert the pressure from torr to atmospheres since the gas constant (R) is given in atm:

P_nitrogen = 38 torr * (1 atm / 760 torr) = 0.05 atm

Now, we can calculate the number of moles of nitrogen using the ideal gas law equation:

n_nitrogen = (P_nitrogen * V) / (R * T)

Since the flask is completely dry, the moles of nitrogen before and after the experiment will be the same.

n_nitrogen(before) = n_nitrogen(after)

(P_nitrogen * V_before) / (R * T_before) = (P_nitrogen * V_after) / (R * T_after)

Since the temperature and gas constant are constant, we can simplify the equation:

V_before = V_after

Now, we can calculate the volume of the flask using the equation:

V = (n_nitrogen * R * T) / P_nitrogen

V = (n_nitrogen * R * T) / P_nitrogen

Using the given values:

n_nitrogen = n_nitrogen(after) = (0.05 atm * V_after) / (R * T)

V = ((0.05 atm * V_after) / (R * T)) * R * T / (0.05 atm)

V = V_after

Substituting the given values:

V = (0.9 g / (28 g/mol)) * (0.0821 L * atm / (mol * K)) * (300 K) / (0.05 atm)

V = 0.9 g * 0.0821 L / 28 g * 300 K / 0.05

V = 24.63 L

Therefore, the volume of the flask is approximately 24.63 L.
Free Test
Community Answer
A mixture of nitrogen and water vapours is admitted to a flask at 760 ...
PH2O=PTotal−PN2


Given : PTotal = 760 torr, PN2 = 722 torr, T=300K


PH2O=760−722=38 torr


PH2O=38/760 atm


nH2O=0.9/18(mass/mol. mass)


Now,

PV=nRT


V=nRT/P


V=0.9×0.821×300×760/18×38


V=24.63 L
Explore Courses for Chemistry exam
A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer?
Question Description
A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Here you can find the meaning of A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A mixture of nitrogen and water vapours is admitted to a flask at 760 torr which contains a sufficient solid drying agent after long time the pressure reached a steady value of 722 torr. If the experiment is done at 27°C and drying agent increases in weight by 0.9 gm, what is the volume of the flask? Neglect any possible vapour of drying agent and volume occupied by drying agent:a)443.34 Lb)246.3 Lc)12.315 Ld)24.63 LCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Chemistry tests.
Explore Courses for Chemistry exam
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