The elevation in boiling point of a solution is $$13.44$$ g of $$CuCl2$$ in $$1$$ kg of water using the following information will be (molecular$$ weight of $$CuCl2$$ $$=134.4$$ and $$Kb=$$ $$0.52$$ $$Km-1).

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Infinity Learn · Accepted Answer

$$CuCl2$$ is an electrolyte, which ionises in solution as follows                                               $$CuCl2$$⇋ $$Cu2++2Cl-$$ At $$t=0$$                   $$1$$ mol            $$0$$           $$0$$ After ionisation    (1-$$α$$)mol$$    α mol   $$2$$α molThus, number of particles after ionisation                 $$=1-$$α$$+$$α$$+2$$α$$=1+2$$α∴van't Hoff factor      $$i=number$$ of particles after ionisationnumber of particles before ionisationor $$i=1+2$$α$$1$$                     on $$100$$% ionisation α$$=1$$      $$=1+2$$×$$11=3The$$ elevation in boiling point $$(when$$ colligative property is $$abnormal)          ∆$$Tb=i$$×Kb×m    $$m-molality$$ of solution  Molality of $$CuCl2$$ solution                               $$=Weight$$ of $$CuCl2$$ (in$$ $$gram)molecular$$ weight of $$CuCl2Weight$$ of water $$(solvent) in kg $$=13.4413.441=0.1$$ m Thus, ∆$$Tb=3$$×$$0.52$$×$$0.1-0.156-0.16oC$$

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