Answer
Step -1 (What is given in problem ?)
condition -1
Partial pressure of ethane = 1 bar
Mass of ethane =6.56 × 10–3 g
Condition -2
Now mass of ethane is 5.00 × 10–2
We have to find the pressure
Condition -2
Now mass of ethane is 5.00 × 10–2
We have to find the pressure
So using Henry’s law
Mass of dissolved gas(m) of directly proportional to its pressure (p)
m= kp here k is
proportionality constant
so plug the values we get
m1= kp1 ……………(1)
m2 = kp2 …………...(2)
Divide equation (2) by (1) equation we get
M2/m1 = p2/p1
Multiply by p1 both sides we get
P2 = m2*p1/m1
Plug the values now we get
P2 = 5.00 × 10–2* 1/6.56 × 10–3 g
P2= 7.62 bar (Answer)
In henry law, p=kx then how it become m=pk?
ReplyDeleteSolubility of mass of gas is directly proportional to pressure by henry law so m=kp
DeleteLet me clear all this folks for ya!!
DeleteMole Frac. = Mass of solute/Molar m of solute/Total moles =Wb/Mb / Wb/Mb+Wa/Ma
You know how its written
So we know here the mass of ehtnae is 6.56 × 10^(-3) and even dividing it by its molar mass will give a even smaller value so we neglect moles of ethane in total moles so now
Xb=Mole frac=Wb/Mb / Wa/Ma
Take Wa as x gm and substitute the value of Kh/x fron first equation by taking Wb as 6.56 × 10^(-3) in second equation and there you have it !! IF ANY doubt please reply !
Yeah ur absolutely correct
ReplyDeleteLet me clear all this folks for ya!!
ReplyDeleteMole Frac. = Mass of solute/Molar m of solute/Total moles =Wb/Mb / Wb/Mb+Wa/Ma
You know how its written
So we know here the mass of ehtnae is 6.56 × 10^(-3) and even dividing it by its molar mass will give a even smaller value so we neglect moles of ethane in total moles so now
Xb=Mole frac=Wb/Mb / Wa/Ma
Take Wa as x gm and substitute the value of Kh/x fron first equation by taking Wb as 6.56 × 10^(-3) in second equation and there you have it !! IF ANY doubt please reply !!