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Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.

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Solution :According to RAOULT's law, `(p^(@)-p_(s))/(p^(@))=x_(2)` (MOLE fraction of the non-volatile solute)
Mole fraction of the volatile liquid, `x_(1)=1-x_(2) or x_(2)=1-x_(1)`
`therefore""(p^(@)-p_(s))/(p^(@))=1-x_(1)""or ""x_(1)=1=(p^(@)-p_(s))/(p^(@))=(p_(s))/(p^(@)).`


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