The Henry-s law constant for the solubility of N2 gas in water at 298K is 1-0 x 105 atm- The mole fraction of N2 is 0-8- The number of moles of N2 from air dissolved in 10 moles of water at 298K and 5 atm pressure is

P_N_2=P_T× mole fraction=5× 0.8=4 From Henry's law P_N_2=K_H× X_N_2 X_N_2=4× 10^ 5 X_N_2=n_N_2n_N_2+n_water n_N_2 lt; lt;n_water X_N_2=n_N_2n_water n_N_2=4× ...

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Standard XI

Chemistry

Henry's Law

The Henry's l...

Question

The Henry's law constant for the solubility of N₂ gas in water at 298K is 1.0 x 10⁵ atm. The mole fraction of N₂ is 0.8. The number of moles of N₂ from air dissolved in 10 moles of water at 298K and 5 atm pressure is

4 x 10^-4

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4 x 10^-5

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5 x 10^-4

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4 x 10^-6

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Solution

The correct option is A
4 x 10^-4

$PN2=PT×mole fraction=5×0.8=4$
From Henry's law
$P_{N_{2}} = K_{H} \times X_{N_{2}}$
$X_{N_{2}} = 4 \times 10^{- 5}$
$X_{N_{2}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{w a t e r}}$
$n_{N_{2}} << n_{w a t e r}$
$X_{N_{2}} = \frac{n_{N_{2}}}{n_{w a t e r}}$
$n_{N_{2}} = 4 \times 10^{- 4}$

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The Henry's law constant for the solubility of N2 gas in water at 298K is 1.0 x 105 atm. The mole fraction of N2 is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298K and 5 atm pressure is

The correct option is A 4 x 10-4 PN2=PT×mole fraction=5×0.8=4 From Henry's law PN2=KH×XN2 XN2=4×10−5 XN2=nN2nN2+nwater nN2<<nwater XN2=nN2nwater nN2=4×10−4

The Henry's law constant for the solubility of N₂ gas in water at 298K is 1.0 x 10⁵ atm. The mole fraction of N₂ is 0.8. The number of moles of N₂ from air dissolved in 10 moles of water at 298K and 5 atm pressure is