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Heptane and octane form an ideal solution. At `373 K`, the vapour pressure of the two liquids are `105.0` kPa and `46.0` kPa, respectively. What will be the vapour pressure, of the mixture of `25 g` of heptane and `35 g ` of octane ? |
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Answer» Correct Answer - `73.43kPa` `p_(hepten e)^(@)=105.2kPa,p_(" octane")^(@)=46.28kPa` `Mw` of heptane `(C_(7)H_(16))=100` `Mw` of octane `(C_(8)H_(18))=114` Moles of heptane `=(26)/(100)=0.26` Moles of octane `=(35)/(114)=0.31` `CHMi_(hepta n e)=(0.26)/(0.26+0.1)=0.456` ` CHMi_(octa n e)=1-0.456=0.544` `p_(hep ta n e)=p_(hepta n e)^(@)xxCHMi_(hep t an e)` `=105.2xx0.456=47.97kPa` `p_(oc t an e)=p_(ocatn e)^(@)xxCHMi_(octa n e)` `=46.8kPaxx0.544=25.46kPa` `p_(t otal)=47.97+25.46=73.43kPa` |
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