Question

The henry law constant for the solubility of N2 gas
in water at 298K is 1.0 x 100000 atm.Then calculate
number of moles of N2 from air dissolved in 10
moles of water at 298K and 5atm pressure?​

Answer 1

GiveN :

• Total Pressure $\sf{(P_T) \: = \: 5 \: atm}$
• Mole fraction of $\sf{N_2 \: = \: 0.8}$
• Henry's Constant $\sf{K_H \: = \: 10^5}$

To FinD :

• Number of moles of Nitrogen

SolutioN :

As we know that :

$\footnotesize\dashrightarrow \boxed{\tt{Partial \: Pressure \: = \: Total \: Pressure \: \times \: Mole \: fraction}} \\ \\ \dashrightarrow \tt{Pressure_P \: = \: 0.8 \: \times \: 5} \\ \\ \dashrightarrow \tt{Pressure_P \: = \: 0.4 \: atm} \\ \\ \underline{\sf{\therefore \: Partial \: Pressure \: is \: 0.4 \: atm}}$

$\rule{150}{0.5}$

Now, use Henry's Law :

$\dashrightarrow \boxed{\tt{Pressure_P \: = \: K_H X}} \\ \\ \dashrightarrow \tt{X \: = \: \dfrac{Pressure_P}{K_H}} \\ \\ \footnotesize \underline{\sf{ \: \: \: \: \: \: \: Where, \: X \: is \: Mole \: Fraction \: And \: X \: = \: \dfrac{n}{10} \: \: \: \: \: \: \:}} \\ \\ \dashrightarrow \tt{\dfrac{n}{10} \: = \: \dfrac{4}{10^5}} \\ \\ \dashrightarrow \tt{\dfrac{n}{10} \: = \: 4 \: \times \: 10^{-5}} \\ \\ \dashrightarrow \tt{n \: = \: 4 \: \times \: 10^{-5} \: \times \: 10^1} \\ \\ \dashrightarrow \tt{n \: = \: 4 \: \times \: 10^{-4} \: moles} \\ \\ \underline{\sf{\therefore \: No. \: of \: moles \: of \: N_2 \: are \: 4 \: \times \: 10^{-4} \: moles}}$