Question

If $N_2$ gas is bubbled through water at $293K$, how many millimoles of $N_2$ gas would dissolve in $300$ moles of water, if $N2$ exerts a partial pressure of 1 bar. Given that Henry's law constant for $N_2$ at $293K$ is $75.00Kbar$.

Answer 1

Let $x$ be the number of mole of $N_2$ dissolved in water.

No. of moles of water $=300$

Given that Henry’s law constant for $N_2$ at $293K$ is $75kbar$.

Partial pressure exerted by $N_2=1bar$

According to Henry's law-

$P_{N_2}=K_H\times x.....\left(1\right)$

Whereas, $x$ is the mole fraction

As we know that,

Mole fraction $=\frac{\text{No. of moles}}{\text{total moles}}$

Therefore,

Mole fraction of $N_2=\frac{x}{(x+300)}$

Now from ${eq}^n\left(1\right)$, we have

$1=(75\times {10}^3)\times \frac{x}{(x+300)}$

$\Rightarrow x+300=75000x$

$\Rightarrow x=\frac{300}{74999}=0.004\text{moles}=4\text{millimoles}$

Hence $4$ millimoles of $N_2$ will dissolve in $300$ moles of water.