9 A mixture of 75 g CHs and 53 g Xe is placed in a container

9. A mixture of 7.5 g CHs and 5.3 g Xe is placed in a container and the total pressure is found to be 0.43 atm. Find the partial pressure of CH4.

atomic weight of Xe= 131.3, atomic weight of C= 12 and that of H=1, molar mass of CH4=12+4*1=16

moles of Xe= mass of Xe/atomic weight= 5.3/131.3=0.0404, moles of methane = mass/molar mass= 7.5/16 =0.47

total moles of mixture= moles of Xe+ moles of CH4= 0.0404+0.47=0.5104

mole fraction = moles/total moles

mole fraction : Xe= 0.0404/0.5104=0.08 and that of CH4= 1-0.08=0.92

partial pressure = mole fraction* total pressure

partial pressures ( atm): Xe= 0.08*0.43=0.0344 and CH4= 0.92*0.43=0.3984