Question

The Henry's law constant for the solubility of $N_2$ gas in water at 298 K is $1.0\times {10}^{5}atm$. The mole fraction of $N_2$ in air is 0.8. The no. of moles of $N_2$ from air dissolved in 10 moles of water of 298 K and 5 atm pressure is:

• A. $2\times {10}^{-5}$
• B. $6\times {10}^{-5}$
• C. $4\times {10}^{-4}$
• D. $9\times {10}^{-6}$

Answer 1

The correct option is C $4\times {10}^{-4}$

$\begin{array}{c}\text{Alc to Henry's law,}P=Kx\\ \begin{array}{c}x\text{is the mole fraction}\\ k\text{is the Henry's law constant}\\ P\text{is partial pressure in atm}\end{array}\end{array}$

$\begin{array}{c}\text{Substituing the values,}\\ \begin{array}{c}c=\frac{4}{1\times {10}^5}=4\times {10}^{-5}\\ \text{no. of moles of}{N}_2\text{in}10\text{moles of water}\\ \frac{n}{10}=4\times {10}^{-5}\\ \Rightarrow n=4\times {10}^{-4}mol\end{array}\end{array}$

Hence ,(C) is correct option.