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# Q5.19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weightof dihydrogen. Calculate the partial pressure of dihydrogen.

10 years ago

let number of moles of dihydrogen and dioxygen be n1 and n2, respectively.

(n1/n2)(1/16)=1/5,(n1/n2)=16/5=3.2

partial pressure of dihydrogen=3.2/(3.2+1)

10 years ago

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/(5*2)=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