the vapour pressure of solvent decreased by 10 mm of mercury when a non volatile solute was added to solvent - Chemistry - Solutions

Question

the vapour pressure of solvent decreased by 10 mm of mercury when a
non volatile solute was added to solvent the mole fraction of the
solute in the solution is 0 2 what should be the mole fraction of
the solvent if the decrease in the vapour pressure is 20 mm of
mercury?

Akanksha Jain · Accepted Answer

Ans is 0 6

You know the equation of relative lowering of vapour pressure which
is given as:
p1ο-p1p1ο=χ2where p1ο=vapour pressure of pure solventp1=vapour pressure of solutionp1ο-p1=decrease in vapour pressureχ2=mole fraction of solute

So, when decrease in vapour pressure is 10 mm Hg, then using above
equation, we get

10 mm Hgp1ο=0
2⇒p1ο=10 mm Hg0 2=50 mm Hg

i e vapour pressure of pure solvent = 50 mm Hg

Now, when decrease in vapour pressure is 20 mm Hg, then using same
equation

2050=χ2⇒χ2=25=0 4

i e mole fraction of solute in solution = 0 4
So, the mole fraction of solvent = 1 - 0 4 = 0 6

the vapour pressure of solvent decreased by 10 mm of mercury when a non volatile solute was added to solvent . the mole fraction of the solute in the solution is 0.2 . what should be the mole fraction of the solvent if the decrease in the vapour pressure is 20 mm of mercury?