1.

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile solute weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of solution 600 mm Hg. What is the molar mass of the solute ?

Answer»

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Solution :`P_(A)^(@)=200" mm Hg, " P_(A)=195 " mm Hg "W_(B)=2g, W_(A)=78 g`
`M_(A)(C_(6)H_(6))=78" g MOL"^(-1)`
ACCORDING to Raoult's Law(solution is ideal or ieal or dilute),
`(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)`
`((200-18=95mm))/(195mm)=((2g))/((M_(B)))xx((78" g mool"^(-1)))/((78g))`
`M_(B)=((2g)xx(78" g mol"^(-1)))/((78g))xx((195mm))/((5mm))=78" g mol"^(-1)`.


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