1.

What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to 4//5^(th) of pure water ?

Answer»

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SOLUTION :`"According to Raoult's Law," (P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=(W_(B)xxM_(A))/(M_(B)xxW_(A))`
`P_(S)=4//5P_(A)^(@),W_(A)=180g, M_(B)=60g mol^(-1),M_(A)=18g mol^(-1)`
`((P_(A)^(@)-0.8P_(A)^(@)))/((0.8P_(A)^(@)))=((W_(B))xx(18g mol^(-1)))/((60g mol^(-1))xx(180g))`
`W_(B)=((0.2)xx(60g mol^(-1))xx(180g))/((0.18)xx(18g mol^(-1)))=150 G`


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