Question

Calculate the mole fraction of O2 and N2 separately dissolved in water at 25°C and 1 atm pressure. Take Henry’s law constant for O2 = 4.34 x 10^4 atm and for N2 = 8.57 x 10^4 atm.

Answer 1

HEYA MATE YOUR ANSWER IS

According to Henry's law, P=K

H

×x

∴x

O

2

=

K

H

P

O

2

=

4.34×10

4

0.2

=4.6×10

−6

Moles of water=

18

1000

=55.5mol

∴x

O

2

=

n

H

2

O

+n

O

2

n

O

2

≈

n

H

2

O

n

O

2

⇒n

O

2

=4.6×10

−6

×55.5=2.55×10

−4

mole

∴ Molarity = 2.55×10

−4

M

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