An aqueous solution of liquid `X` ( mol weight `56`) `28%` by weight has a vapour pressure `150mm`. Find the vapour pressure of `X` if vapour pressure of water is `155 mm of Hg`.

Explanation

According to Raoult's law of liquid mixutes,
`P_(M)=P'_(A)+P'_(B)`
`:. P_(M)=P_(A)^(@)xx{((w_(A))/(m_(A)))/(w_(A)/m_(A)+(w_(B))/(m_(B)))}+P_(B)^(@)xx{((w_(B))/(m_(B)))/(w_(A)/m_(A)+(w_(B))/(m_(B)))}`
Given, `w_(A)= 28 g, w_(H_(2)O)= 72 g, P_(A)^(@)=?`
`P_(overset(@)H2O)= 155, m_(A)= 56 g, m_(H_(2)O)= 18 g`,
and `P_(M)= 150 mm`
`:. 150=P_(A)^(@)xx{((28)/(56))/((28)/(56)+(72)/(18))}+155xx{((72)/(18))/((28)/(56))+(72)/(18)}`
`150=P_(A)^(@)xx(1)/(2)xx(2)/(9)+155xx4xx(2)/(9)`
`:. P_(A)^(@)= 110 mm`