The value of Van't Hoff factor for aqueous KCL sol is close to 2, while the value of 'i' for ethanoic acid in benzene is nearly 0.5. Give reason?
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Solution
Because vant hoff factor=final moles/initial moles kcl->k+ +cl- initial moles =1 final moles=2 so vant hoff factor=2/1=2 now ethanoic acid in benzene dimerises means 2 molecules become 1 hence initial moles=2 and final moles=1 vant hoff factor=1/2=0.5 I think this helps