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The vapour pressure of chlorobenzene and water at different temperature are `{:(t//^(@)C,90,100,110),(P^(@)(phiCI)//mmHg,204,289,402),(P^(@)(H_(2)O)//mmHg,526,760,1075):}` (A)At what pressure will `phiCl` steam-disillation at `90^(@)C`? (B)At what temperature will `phiCl` steam-dissillation under a total pressure of `800 mm Hg`? (C) How many grams of steam are required for distillation of `10.2g` of `phiCl` (i)at `90^(@)C` and (ii) under `800` torr total pressure? |
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Answer» Correct Answer - (A)`730mm Hg.` (B)`approx92^(@)C`, (C) `4.13gm,4.14gm` (A) At `90^(@)C`, Pressure at which `phiCl` steam `-` distil `=P^(@)(phiCl)+P^(@)(H_(2)O)` `=204+526=730 mm` of `Hg` (B) At `90^(@)C` Total `P_(T)=730mm Hg` At `100^(@)C` total `P_(T)=289+760=1049 mm Hg` so for `800 mm Hg`, temperature will lie in between `90^(@)-100^(@)C` Using extrapolation method. Temperature `=90+((800-730))/((1049-730))xx(100-90)=92.19^(@)C`. (C) (i) `(P_(phiCl)^(@))/(P_(H_(2)O)^(@))=(n_(phicl))/(n_(H_(2)O))` `(204)/(526)=(10xx18)/(112.5xxm)` `m=4.13gm` (ii) At `800` tprr `T=92^(@)C` For `phiCl,(p-204)/(92-90)=(289-204)/(100-90)` `P=572.8` For `H_(2)O,(p-526)/(92-90)=(760-526)/(100-90)` `P=572.8` `(10xx18)/(112.5xxm)=(221)/(572.8 rArrm=4.14gm` |
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