At 298 K 1.0 g of a non-colatile solute is dissoved in 100 g of acetone (mol mass=58). The vapour pressure of the solution at this temperature is found to be 192.5 mm Hg. Calculate themolar mass of the solute. The vapour pressure of pure acetone at 298 K is found to be 195 mm Hg.

Explanation

Mass of solute, `W_(B)=1.0 g`
Mass of solvent, `W_(A)("acetone")=100 g`
Molar mass of solvent `M_(A)("acetone")=58g mol^(-1)`
Vapour pressure of oslution, `P_(S)=192.5 mm Hg`
Vapour pressure of pure solvent (acentone), `P_(A)^(@)=195 mm Hg`
According to Raoult's Law, `(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(A)xxM_(A)/W_(A)`
`((195mm-192.5mm))/((192.5mm))=((1.0g)xx(58gmol^(-1)))/(M_(B)xx(100g))`
`0.013=((1.0g)xx(58gmol^(-1)))/(M_(B)xx(100g))`
`M_(B)=((1.0g)xx(58g mol^(-1)))/(0.013xx(100g))=44.52 g mol^(-1)`