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The normal boiling point of watr is 373 K (at 760 mm), vapour pressure of water at 298 K is 23 mm. If entyalpyof vaporization is 40.656 kj//mol.the boiling point of water at 23 mm pressure will be : |
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Answer» <P>250 K `"log" (P_(2))/(P_(1))=(DeltaH_(v))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `rArr"log" (760)/(23)=(40656)/(2.303xx8.314)[(373-T_(1))/(373xxT_(1))]` `:.T_(1)=294.4K`. Hence, (B) is the correct answer. |
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