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Solubility of a gas in a liquid

The Henry's law constant for the solubility of {N}_{2} gas in

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The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 1 0^{5} atm$ . The mole fraction of $N_{2}$ in air is $0.8$ . The number of moles of $N_{2}$ from air dissolved in 10 moles of water of $298 K$ and $5 atm$ pressure is

4.0 \times 1 0^{- 4}

4.0 \times 1 0^{- 5}

5.0 \times 1 0^{- 4}

4.0 \times 1 0^{- 6}

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Updated on: Feb 17, 2024

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New Pattern JEE Problems Chemistry for JEE Main and Advance( Master Problem Package) (Arihant)

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The Henry's law constant for the solubility of

N_{2}

gas in water at

298 K

1.0 \times 1 0^{5} atm

. The mole fraction of

N_{2}

in air is

0.8

. The number of moles of

N_{2}

from air dissolved in 10 moles of water of

298 K

and

5 atm

pressure is

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

(a)

4.0 \times 1 0^{- 4}

Total pressure is

5 atm

. The mole fraction of Nitrogen is

0.8

. Hence, the partial pressure of Nitrogen

=

mole fraction of Nitrogen

\times

Total pressure

=

0.8 \times 5 = 4

atm According to Henry's law,

P = K x

x

is the mole fraction

K

is the henry's law constant

P

is the pressure in atm. Substitute values in the above expression.

x = \frac{4 atm}{1 \times 1 0 ^{5} atm} = 4 \times 1 0^{- 5} = \frac{n}{n + 10}

where

n

is no. of mole of nitrogen

n = 4.0 \times 1 0^{- 4} mol

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Question Text	The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 1 0^{5} atm$ . The mole fraction of $N_{2}$ in air is $0.8$ . The number of moles of $N_{2}$ from air dissolved in 10 moles of water of $298 K$ and $5 atm$ pressure is
Updated On	Feb 17, 2024
Topic	Solutions
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1 Video solution: 27
Upvotes	2892
Avg. Video Duration	5 min

The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 1 0^{5} atm$ . The mole fraction of $N_{2}$ in air is $0.8$ . The number of moles of $N_{2}$ from air dissolved in 10 moles of water of $298 K$ and $5 atm$ pressure is

Text solutionVerified

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Text solutionVerified

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The Henry's law constant for the solubility of $N_{2}$ gas in water at $298 K$ is $1.0 \times 1 0^{5} atm$ . The mole fraction of $N_{2}$ in air is $0.8$ . The number of moles of $N_{2}$ from air dissolved in 10 moles of water of $298 K$ and $5 atm$ pressure is