`X_(A)` and `X_(B)` are the mole fraction of A and B respectively in liquid phase `y_(A)` and `y_(B)` are the mole fraction of A and B respective in vapour phase. Find out the slope of straight line if a graph is plotted `(1)/(y_(A))` along Y-axis against `(1)/(x_(A))` along X-axis gives straight line `[p_(A)^(@)` and `p_(B)^(@)` are vapour pressure of pure components A and B].

Answer

A

Explanation

`y_(A)=(p_(A))/(p_(A)-p_(B))`
`=(x_(A)p_(A)^(@))/(x_(A)p_(A)^(@)+x_(B)p_(B)^(@))`
`=(x_(A)p_(A)^(@))/(x_(A)p_(A)^(@)+(1-x_(A))p_(B)^(@))`
`=(x_(A)p_(A)^(@))/(x_(A)(p_(A)^(@)-p_(B)^(@))+p_(B)^(@))`
`(1)/(y_(A))=(x_(A)(p_(A)^(@)-p_(B)^(@))+p_(B)^(@))/(x_(A)p_(A)^(@))`
`=(p_(A)^(@)-p_(B)^(@))/(p_(A)^(@))+(p_(B)^(@))/(p_(A)^(@)(x_(A)))`
On comparing with general straight line equation `y = mx + c`, we get
`m="slope"=(p_(B)^(@))/(p_(A)^(@))`
`c=" intercept"=(p_(A)^(@)-p_(B)^(@))/(p_(A)^(@))`