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.

The vapour pressure of a solution containing 108.24 g of a non-volatil

1.	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» Solution :`"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).`