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The vapour pressure of pure watr at 30^(...

The vapour pressure of pure watr at `30^(@)C is 31.80` mm of Hg. How many grams of urea (molecular mass=60) should be dissolved in 1000 g of water to lower the vapour pressure by 0.25 mm of Hg?

Text Solution

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The correct Answer is:
62.2 g
`P_(A)^(@)=31.80 mm , P_(A)^(@)-P_(S)=31.80-0.25=31.55mm`
`W_(A)=1000g, M_(B)=60" g mol"^(-1), M_(A)=18" g mol"^(-1)`
According to Raoult's Law,
`(P_(A)^(@)-P_(S))/(P_(S))=x_(B)=(W_(B)xx(18" g mol"^(-1)))/((60" g moll"^(-1))xx(1000g))`
`W_(B)=(0.25)/(31.80)xx((60" g mol"^(-1))xx(1000g))/((18" g mol"^(-1)))=26.2g`
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