$5.5 m g$ of nitrogen gas dissolves in $180 g$ of water at $273 K$ and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in $180 g$ of water at $5 a t m$ nitrogen pressure is approximately :

1 \times 10^{- 5}

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1 \times 10^{- 4}

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1 \times 10^{- 6}

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1 \times 10^{- 3}

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Solution

The correct option is C $1 \times 10^{- 4}$
According to Henry's law:

$P_{g a s} = K_{H} . X_{g a s}$

In the above case we can write:

$\frac{P_{1}}{P_{2}} = \frac{X_{1}}{X_{2}}$

$\frac{1}{5} = \frac{\frac{(5.5 \times 10^{- 3} / 28)}{(5.5 \times 10^{- 3} / 28) + (180 / 18)}}{X_{2}}$

$\Rightarrow X_{2} = 1 \times 10^{- 4}$

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5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately :

The correct option is C 1×10−4According to Henry's law:Pgas=KH.XgasIn the above case we can write:P1P2=X1X215=(5.5×10−3/28)(5.5×10−3/28)+(180/18)X2⇒X2=1×10−4

$5.5 m g$ of nitrogen gas dissolves in $180 g$ of water at $273 K$ and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in $180 g$ of water at $5 a t m$ nitrogen pressure is approximately :