Submit at 25.0 c, a 10.00 l vessel is filled with 5.00 atm of gas a and 7.89 atm of gas b. what is the mole fraction of gas b?

the mole fraction of Gas B is xB= 0.612 (61.2%)

Explanation:

Assuming ideal gas behaviour of A and B, then

pA*V=nA*R*T

pB*V=nB*R*T

where

V= volume = 10 L

T= temperature= 25°C= 298 K

pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm

R= ideal gas constant = 0.082 atm*L/(mol*K)

therefore

nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole

nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole

therefore the total number of moles is

n = nA +nB= 2.04 mole +  3.22 mole = 5.26 mole

the mole fraction of Gas B is then

xB= nB/n= 3.22 mole/5.26 mole = 0.612

xB= 0.612

Note

another way to obtain it is through Dalton's law

P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612

the answer would be c

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.