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»

<P>10 g
11.4 g
9.8 g
12.8 g

Solution :`(p_(A)^(@)-p_(A))/(p_(A)^(@))=(W_(B)M_(A))/(W_(A)M_(B))`
Lowering in VAPOUR pressure is 20%
`:.(p_(A)^(@)-p_(A))/(p_(A)^(@))=(20)/(100)=0.2`
`W_(B)=?, W_(A)=114g`
`M_(B)=40g, M_(A)=114g`
`:.0.2=(W_(B)xx114)/(20xx114)` or `W_(B)=8G`.


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