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Dalton's law partial pressure

A gaseous mixture was prepared by taking equal mole of {C O} a

Question

Medium

Solving time: 3 mins

A gaseous mixture was prepared by taking equal mole of $CO$ and $N_{2}$ . If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is:

1 atm

0.5 atm

0.8 atm

0.9 atm

Views: 5,304 students

Updated on: Apr 25, 2023

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Henry Discussed

A gaseous mixture was prepared by taking equal mole of

CO

and

N_{2}

. If the total pressure of the mixture was found

1

atmosphere, the partial pressure of the nitrogen

(N_{2})

in the mixture is:

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

According to Dalton's Law of partial pressure.
The partial pressure of gas

\propto

Mole fraction of gas (

X

gas)

X_{CO} = \frac{n _{CO}}{n _{CO} + n _{N_{2}}} = \frac{n _{CO}}{2 n _{CO}} [∵ n_{CO} = n_{N_{2}}] \Rightarrow X_{CO} = \frac{1}{2} = 0.5

Total pressure of the mixture was found

1

atmosphere,
Thus, Partial pressure of

N_{2} = X_{N_{2}} \cdot P_{T} = [1 - X_{CO}] \cdot P_{T} = [1 - 0.5] \times [1] = 0.5 atm .

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Question Text	A gaseous mixture was prepared by taking equal mole of $CO$ and $N_{2}$ . If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is:
Updated On	Apr 25, 2023
Topic	States of Matter
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 8
Upvotes	875
Avg. Video Duration	10 min

A gaseous mixture was prepared by taking equal mole of $CO$ and $N_{2}$ . If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is:

Text solutionVerified

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A gaseous mixture was prepared by taking equal mole of $CO$ and $N_{2}$ . If the total pressure of the mixture was found $1$ atmosphere, the partial pressure of the nitrogen $(N_{2})$ in the mixture is: