The Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at 298 K is $1.0 \times 10^5\mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8. The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Solutions

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 at 298 K and 5 atm pressure is

Answer: (A)

According to Henry's law:

P = K_{H} χ_{N_{2}}

The partial pressure of

N_{2}

in air at 5 atm total pressure:

P_{N_{2}} = 0.8 \times 5 = 4 atm

Using Henry's law:

4 = 1.0 \times 1 0^{5} \times χ_{N_{2}}

χ_{N_{2}} = 4 \times 1 0^{- 5}

This is the mole fraction of

N_{2}

in the solution containing 10 moles of water:

χ_{N_{2}} = \frac{n _{N_{2}}}{n _{N_{2}} + 10}

Since

n_{N_{2}}

is very small compared to 10:

4 \times 1 0^{- 5} \approx \frac{n _{N_{2}}}{10}

n_{N_{2}} = 4 \times 1 0^{- 4} moles

Therefore, the answer is **(A)

4.0 \times 1 0^{- 4}

**.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option