`5g` of compound `A` was dissolved in `100g` of water at `303K`. The vapour pressure of the solution was found to be `4.16` kilopascal. If the vapour pressure of pure water is `4.24 kPa` at this temperature, what is the molecular mass of `A`?

Explanation

Given Mass of solute `(A)=5g`
Mass of solvent (water)`=100g`
vapour pressure of the solution`=4.16` killopasical
vapour pressure of solvent `=4.24kPa`
Asked molecular mass of solute `=?`
Formulae: `(P_(A)^(0)-P_(A))/(P_(A)^(0))=x_(B)=(W_(B)/(M_(B)))/(W_(A)/M_(A)+W_(B)/(M_(B)))`
Explanation : `P_(A)^(0)=` vapour pressure of pure solvent
`P_(A)=` vapour pressure of solution
`X_(B)=` Mole fraction of solute
`W_(B)=` Molecular weight solute
`W_(A)=` Mass of solvent
`M_(A)=` Molecular weight of solvent
Substitution `&` Calculation
`(4.21-4.16)/(4.24)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08)/(4.23)=(5)/(M_(B)/(100/(18)+(5)/(M_(B))))rArr(0.08xxM_(B))/(M_(B))+0.08xx(100)/(M_(B))xx4.24`
`rArr(0.40)/(M_(B))+(8)/(18)=(21.20)/(M_(B))rArr(20.80)/(M_(B))=(8)/(18)`
`rArr M_(B)=(20.80xx18)/(8)=46.8g mol^(-1)`.