A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen ?

Hello Student
Let the mass of the misture = M
Then M/5 (20%) is constituted by the dihydrogen gas and the remaining 4M/5 (80%) is constituted by dioxgen gas.
M/5 Mass of dihygrogen contains M/(5xxx2)=M/10 mols of dihydrogen and 4M/5 mass of dioxygen will contain 4M/(5*32)=M/40 Mols of dioxygen. As sucj /Mol Fraction of dihydrogen in the mixture=(M/10)/((M/40)+(M/10))=(M/10)/(M/8)=(4/5)
As Such The partial pressures is mol fraction* total pressures = (4/5)*1 bar = 0.8 bar