Assuming ideal behavior, what mass of naphthalence `(C_(10)H_(8))` would have to be dissolved in `200 g` of octane `(C_(8)H_(18))` to lower the vapor pressure of pure octane by `20%`

Options

A. `56 g`
B. `47 g`
C. `69 g`
D. `35 g`

Answer

3

Explanation

If the vapour pressure of pure octane is `P_(oct)^(0)`, then the desired vapor pressure above the solution is `80%` of `P_(oct)^(0)` or `0.80 P_(oct)^(0)`. According to Raoult's law:
`P_(oct)=P_(oct)^(0)chi_(oct)`
`0.80 P_(oct)^(0)=P_(oct)^(0)chi_(oct)`
`chi_(oct)=0.80`
By definition
`chi_(oct)=n_(oct)/(n_(oct)+n_(C_(10)H_(8))`
The molecular mass of octane is 114, and the molecular mass of naphthalence is 128. Thus:
`0.80=((200 g " octane")/(114 g " octane"//1 mol " octane"))/((200" g octane")/((114 g " octane")/(1 mol " octane"))+(200 g " octane")/((114 g " octane")/(1 mol " octane"))`
`mass_(C_(10)H_(8))=56 g`