1.

The vapour pressure of pure benzene at a certain temperature is 200 mm Hg. A non-volatile solute weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of solution is 600 mm Hg. What is the molar mass of the solute ?

Answer»


Solution :`P_(A)^(@)=640" mm HG, "P_(S)=600" mm Hg, " W_(B)=2.175h, W_(A)=39.0 g.`
`M_(A)(C_(6)H_(6))=78" g mol"^(-1)`
According to Raoult's Law (solution is IDEAL or dilute),
`(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)`
`((640-600)mm)/(600mm)=((2.175g))/((M_(B)))xx((78" g mol"^(-1)))/((39.0 g))`
`M_(B)=((2.175 g)xx(78" g mol"^(-1)))/((39.0 g))xx((78" g mol"^(-1)))/((40 mm))=65.25" g mol"^(-1).`


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