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8/19
Let take the solution of 100 gram
Oxygen =36gram
Acetaldehyde = 64gram
noxygen=36/18=2mole=n1
acetaldehyde=64/44=1.5=n2
Mole fraction of acetaldehyde= n2/(n1+n2)=0.41
Let us consider 100gm of solution
So the solution contains 36gm water and 64gm acetaldehyde(C2H4O)
So no. of moles of water = 36/18 = 2moles
& no. of moles of acetaldehyde = 64/44 = 16/11 moles
Now mole fraction of water = 2/(2+16/11) = 2*11/38 = 11/19
& mole fraction of acetaldehyde = 16*11/11*38 = 8/19
acetaldehyde mole fraction is 0.64
mole fraction of CH3COOH is 0.347
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