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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
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B
4×105
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C
5×104
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D
4×106
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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 fraction
K is the henry's law constant
P 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

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