The vapour pressure of ethanol and methanol are 44.5 and 88.7 mm of Hg at 298 K. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour phase.

Answer

`p_(("solution"))= 66.16 m m Hg`

Explanation

`n_("ethanol ") = ( 60 g)/( 46 g mol ^(-1)) = 1 . 30 ` mol
`n_(methanol ") = (40 g)/( 46 g mol ^(-1)) = 1 . 25 ` mol
(Molar mass o f methanol , `CH_(3) OH = 32 g mol^(-1)` Molar mass of ethanol , `C_(2 H_(5) OH = 46b g mol ^(-1)`
` x_("ethanol ") = (1 . 30)/( . 30 + 1 . 25) = (1 . 30)/( 2 . 55) = 0 . 51 `
`x_(" methanol ") = 1 - x ethanol = 1 - 0 . 51 = 0 . 49`
`p_("solution ") = p_("methanol ") + p_("ehtanol ")`
`= p_("methanol ")^(@) xx x_("methanol") + p_("ethanol")^(@) xx x_("ethanol ")`
` = (88 . 7 xx 0 . 49 + 44 . 5 xx 0 . 51 ) m m Hg `
`= (43 . 46 + 22 . 69 ) m m Hg`
`= 66 . 15 m m Hg`
`y_("methanol ") = (p_("methanol"))/(p_("solution ")) = (43 . 46)/( 66 . 15) = 0 . 657`
`y_("ethanol ") = 1 - y _("methanol") = 1 - 0657`
= 0 . 343