Answer: Missing data related to question is attached below
If the KH,CH4 = 1.42 * 10^-3 mol/L atm
answer:
0.88 atm
Step-by-step explanation:
P1 = 20 psi
aqueous concentration of methane in closed bottle = 20 ppm
20 ppm = 20 / 1000 ( gram / Kg ) = 0.020 gram/kg = 0.020 g/liter
molar mass of methane = 16 gram/mol
next : convert mass conc to molar conc
Cm = 0.020 / 16
Cm = 1.25 * 10^-3 mol/L
Given that KH,CH4 = 1.42 * 10-3 mol/L atm
Applying equilibrium relationship
Cm = ( KH,CH4 ) ( partial pressure of methane )
hence partial pressure of methane
= Cm / ( KH,CH4 ) = 1.25 * 10^3 / (1.42 * 10^-3 ) = 0.88 atm