100g of liquid A (molar mass 140g `"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

Number of moles of A,
`n_(A)=(100g)/(140" g mol"^(-1)) = 0.714` mol
Number of moles of B,
`n_(B)=(1000g)/(180"g mol"^(-1))=5.555` mol
Total number of moles of A and B
` =0.714 +5.555 =6.269 `
Mole fraction of A, `x_(A)= (0.714)/(6.269)=0.114`
Mole fraction B,`x_(B)=(5.555)/(6.269)=0.886`
Vapour pressure of pure A, `p_(A)^(0)=`?
Vapour pressure of pure B, `p_(B)^(0)=500` torr
`P_("total") = 475` torr
According to Raoult.s law :
`P_("total") = p_(A)^(0) .x_(A)+p_(B)^(0).x_(B)`
475 torr `=p_(A)^(0) xx 0.114 +500 xx 0.886 = p_(A)^(@)xx0.114 +443` torr
`:.p_(A)^(0) xx 0.114 =(475-443)`torr = 32 torr
`:.` Vapour pressure of A in solution, `p_(A)=32` torr
Vapour pressure of pure A, `p_(A)^(0)=(32)/(0.114)=280` torr