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If alpha for alcl3 is 50% …

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If alpha for alcl3 is 50% find the value of van't hoff factor
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AlCl3---- Al + 3Cl here @= alpha ---degree of dissociation @= 0.5 since. Vant hoff factor = observed number of moles divvided by normal number of moles Initial number ofmoles---- 1,0,0. Final number of moles ----alcl3 = 1-@. Al= @. 3cl= 3@. i=[1-@+@+3@]. i=[1+3@]. i= 1+(3*0.5). i= 2.5
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