At 298 K and 1 atm pressure - solubility of N2 in water was found to be 6-8-10-4 mol-L - if the partial pressure of N2 is 0-78 atm - then what is the concentration of N2 dissolved in water under atmospheric conditions

Henry's #160;Law #160;: #160;p=KHCp=partial #160;pressureKH=Henry's #160;ConstantC #160;= #160;ConcentrationSo, #160;Solubility #160;of #160 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Solubility

at 298 K and ...

Question

at 298 K and 1 atm pressure , solubility of N₂ in water was found to be 6.8*10^-4 mol/L . if the partial pressure of N₂ is 0.78 atm . then what is the concentration of N₂ dissolved in water under atmospheric conditions

Open in App

Solution

$H e n r y' s L a w : p = K_{H} C p = p a r t i a l p r e s s u r e K_{H} = H e n r y' s C o n s t a n t C = C o n c e n t r a t i o n S o, S o l u b i l i t y o f N_{2} a t 1 a t m = 6.8 \times 10^{- 4} m o l / L p = 1 a t m S o, K_{H} = \frac{p}{C} = \frac{1}{6.8 \times 10^{- 4}} = 1.47 \times 10^{3} a t m / m o l W h e n t h e p a r t i a l p r e s s u r e = 0.78 a t m C = \frac{p}{K_{H}} = \frac{0.78}{1.47 \times 10^{3}} = 0.53 \times 10^{- 3} = 5.3 \times 10^{- 4} m o l / L$

Suggest Corrections

Similar questions

Q. The Henry's law constant for the solubility of

N_{2}

gas in water at

298 K

1 \times 10^{5} atm

. The mole fraction of

N_{2}

in air is

0.8

. Then the number of moles of

N_{2}

dissolved in

10

moles of water at

298 K

and

5 atm

pressure is:

Q. Henry's law constant for the solubility of

N_{2}

gas in water at

298

K is

1.0 \times 10^{5}

atm. The mole fraction of

N_{2}

in air is

0.6

. The number of moles of

N_{2}

from air dissolved in

10

moles of water at

298

K and

5

atm pressure is:

Q. The Henry's law constant for the solubility of

N_{2}

gas in water at 298 K is

1 \times 10^{5}

atm. The mole fraction of

N_{2}

in air is 0.8. The number of moles of

N_{2}

from a dissolved in 100 moles of water at 298 K and 5 atm pressure is:

Q. The Henry's law constant for the solubility of

N_{2}

gas in water at

298 K

1 \times 10^{5}

atm. The mole fraction of

N_{2}

in air is

0.8

. Calculate the number of moles of

N_{2}

dissolved in

10 moles

of water at

298 K

and

5 atm

Q. The Henry's law constant for the solubility of

N_{2}

gas in water at

298 K

1.0 \times 10^{5} atm

. The mole fraction of

N_{2}

in air is

0.8

. The number of moles of

N_{2}

dissolved in

10

moles of water at

298 K

and

5 atm

x \times 10^{- 4}

. Find the value of

x

Join BYJU'S Learning Program

at 298 K and 1 atm pressure , solubility of N2 in water was found to be 6.8*10-4 mol/L . if the partial pressure of N2 is 0.78 atm . then what is the concentration of N2 dissolved in water under atmospheric conditions

Henry's Law : p=KHCp=partial pressureKH=Henry's ConstantC = ConcentrationSo, Solubility of N2 at 1atm = 6.8×10-4mol/Lp=1 atmSo, KH=pC=16.8×10-4=1.47×103 atm/molWhen the partial pressure = 0.78 atmC = pKH=0.781.47×103=0.53×10-3=5.3×10-4 mol/L

at 298 K and 1 atm pressure , solubility of N₂ in water was found to be 6.8*10^-4 mol/L . if the partial pressure of N₂ is 0.78 atm . then what is the concentration of N₂ dissolved in water under atmospheric conditions