1.

The mass of a non-volatile solute of molar mass 40 g mol^(-1) that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is

Answer»

10 g
11.4 g
9.8 g
12.8 g

Solution :Using Raoult's Law,
`(P_(o)-P_(s))/(P_(s))=(w)/(m)xx(M)/(W)`
If `P_(o)=100 mm`, then `P_(s)=80mm`
`(100-80)/(80)=(w)/(114)xx(114)/(40) "" THEREFORE "" w=10 g`


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