Question

If \(N_2\) gas is bubbled through water at \(293~K\), how many millimoles of \(N_2\) gas would dissolve in \(1~litre\) of water? Assume that \(N_2\) exerts a partial pressure of \(0.987\) bar. Given that Henry’s law constant for \(N_2\) at \(293~K\) is \(76.48~kbar\).

Answer 1

Mole fraction of \(N_2\) gas

Given:

Partial pressure of \(N_2=0.987~bar\) and Henry's law constant for \(N_2=76.48~kbar=76.48\times 10^{3} bar\).

According to Henry's law,

Partial pressure \(p=K_H\times \text{moles fraction(x)}\)

So, \(x_{N_2}=\dfrac{P_{N_2}}{K_H}=\dfrac{0.987~bar}{76.48\times 10^3bar}\)

\(=1.29\times 10^{-5}\)

Moles of water

Given: Volume of water \(=1~L\)

We know, \(\text{density(d)}=\dfrac{\text{mass}}{\text{volume}}\)

We also know,

Density \((d)\) of water \(=1kg/L\)
Hence, mass of water \(=1~kg=1000~g\)
Molar mass of water\((H_2O)=2\times 1+16=18gmol^{-1}\)

\(\text{moles of water}=\dfrac{\text{mass of water}}{\text{molar mass of water}}\)
\(=\dfrac{1000~g}{18~g~mol^{-1}}=55.55~mol\)

Millimoles of \(N_2\) gas

Let the mole of \(N_2\) gas be \(n\).
Mole fraction \((x_{N_2})=\dfrac{\text{moles of}~N_2}{\text{moles of}~N_2+\text{moles of}~H_2O}\)

By putting the values, we get

\(1.29\times 10^{-5}=\dfrac{n~mol}{n~mol+55.5~mol}\)

\(n< < 55.5\)

\(1.29\times 10^{-5}=\dfrac{n}{55.5}\)
Thus, \(n=1.29\times 10^{-5}\times 55.5~mol\)
\(=7.16\times 10^{-4}~mol~or~0.716~mmol\)