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Vapour pressure of a solution is less than vapour pressure in pure state . why?

Ajay , 7 Years ago
Grade 12
anser 1 Answers
Vikas TU

Last Activity: 7 Years ago

From Raoult's law; as per this, when a nonvolatile solute is broken down in a dissolvable, it brings down the vapor weight of the dissolvable (vapor weight bringing down) on account of the associations amongst dissolvable and solute particles. 
Right now, dissolvable to solute attractions are what will make less dissolvable particles vanish, customarily in light of the fact that there are just less dissolvable atoms at the surface, and on the grounds that dissolvable particles now require more vitality to escape into gas frame. 
Raoult's law expresses that for a perfect arrangement, the halfway weight of a segment show in that arrangement is equivalent to the mole division of that part times its vapor weight when unadulterated.

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