The vapour pressure of a solution containing 108.24 g of a non-volatile solute dessolved in 1000 g of water at 293 K is 17.354 mm Hg. The vapour pressure of water at 293 K is 17.54 mm Hg. Assuming the solution to be ideal, Calculate the molar mass of the solute.

Answer

181.78 h `mol^(-1)`

Explanation

`"Mass of solution "(W_(B))=108.24 g`
`"Mass of solvent "(W_(A))=1000 g`
`"Molar mass of solvent "(M_(A))=18" g mol"^(-1)`
`"Vapour pressure of solvent "(P_(A)^(@))=17.54 mm.`
`"Vapour pressure of solution "(P_(S))=17.354 `
According to Raoult's Law (Solution is non-ideal):
`(P_(A)^(@)-P_(S))/(P_(S))=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)`
`((17.54-17.354)mm)/((17.354 mm))=((108.24g))/(M_(B))xx((17.354mm))/((0.186mm))=181.78" g mol"^(-1)
`M_(B)=((108.24g)xx(18" g mol"^(-1)))/((1000g))xx((17.354mm))/((0.186mm))=181.79" g mol"^(-1).`