Question

# The Henry's law constant for the solubility of $$N_2$$ gas in water at $$298K$$ is $$1.0\times 105\, atm$$. 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 $$298K$$ and $$5 \, atm$$ pressure is

A
4×104
B
4×105
C
5×104
D
4×106

Solution

## The correct option is B $$4 \times 10^{-4}$$The total pressure is 5 atm. The mole fraction of Nitrogen is 0.8. Hence, the partial pressure of Nitrogen = mole fraction of Nitrogen × Total pressure = $$0.8 \times 5=4 \ atm$$According to Henry's law, $$P=Kx$$x is the mole fractionK is the henry's law constantP is the pressure in atm.Substitute values in the above expression.$$x=\dfrac {4 \ atm} {1 \times 10^{5}atm}=4 \times 10^{-5}= \dfrac {n} {n+10}$$where n is no. of mole of nitrogen$$n=4.0 \times 10^{-4}mol$$So, the correct option is $$A$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More