The vapour pressure of ethanol and methanol ate `44.5 mm Hg` and `88.7 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60g` of ethanol and `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.

Explanation

No. of moles of `C_(2)H_(5)OH=60/46=1.304, "No. of mole of "CH_(3)OH=40/32=1.25`
` 'X_(A)' "of ethyl alcohol"=1.304/(1.304+1.25)=0.5107`
`'X_(B)' "of methyl alcohol"=1.25/(1.304+1.25)=0.4893`
Partial pressure of ethyl alcohol`=X_(A).P_(A)^(0)=0.5107xx44.5=22.73mm" Hg"`
Partial pressure of methyl alcohol `=X_(B).P_(B)^(0)=0.4893xx88.9=43.40mm" Hg"`
Total vapour pressure of solution`=22.73+43.40=66.13mm" Hg"`
Mole fraction of methyl alochol in the vapour `=("Partial pressure of " CH_(3)OH)/"Total vapour pressure"=43.40/66.13=0.6563`