`100g` of liquid `A(` molar mass `140 g mol ^(-1))` was dissolved in `1000g ` of liquid `B(` molar mass `180g mol^(-1))`. The vapour pressure of pure liquid `B` was found to be `500` torr. Calculate the vapour pressure of pure liquid `A` and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr

Explanation

No. of moles of solute, `n_(2)=(100)/(140)=(5)/(7)` mole
No. of moles of solvent `n_(1)=(1000)/(180)=(50)/(9)` mole
Mole fraction of solute
`x_(2)=(n_(2))/(n_(1)+n_(2))=(5//7)/(5//7+50//9)=0*114`
Mole fraction of solvent `x_(1)=(1-x_(2))=(1-0*114)`
`=0*886`
According to Raoult's law
`P_(A)=x_(A)P_(A)^(@)=0*114xx10P_(A)^(@)`
`P_(B)=x_(B)P_(B)^(@)=0*886xx500=443` torr
`P_("total")=P_(A)+P_(B)`
`475=0*114P_(A)^(@)+443`
`P_(A)^(@)=(475-443)/(0*114)=280*7` torr
`threforeP_(A)=0*114xx280*7=32` torr.