At `20^(@)C`, The vapour pressure of liquid A is 22 mm of Hg and that of pure liquid B is 75 mm of Hg. What is the composition of these somponents in the solution that has the vapour presure of 48.5 mm of Hg at this tempreture ? (Assuming ideal solution behaviour).

Answer

Same no. of moles

Explanation

According to Raoult's Law.
`P_(A)=P_(A)^(@)+P_(A)=P_(A)^(@)x_(A)orP_(A)=(22mm)x_(A)`
`P_(B)=P_(B)^(@)x_(B)orP_(B)=(22 mm)x_(B)=(75 mm)(1-x_(A))`
According to availble data :
`P_(A)+P_(B)=48.55 mm`
`therefore (22mm)x_(A)+(75 mm)(1-x_(A))=48.55 mm`
`therefore (-53 mm)x_(A)=-26.45 mm`
or `x_(A)((-26.45mm)/(-53mm))=0.5`
Thus `x_(A)=0.5" and "x_(B)=(1-0.5)=0.5`
`therefore "Two liquids have equal moles of A and B".`