We know that, For NaCl i = 2 For BaCl2 = 3 ∆TF (NaCl) / ∆TF (BaCl2 ) = 2 / 3 Hence, ∆TF (BaCl2 ) = 3 x 2 / 2 = 3 TF = – 3◦C
We know that, For NaCl i = 2 For BaCl2 = 3 ∆TF (NaCl) / ∆TF (BaCl2 ) = 2 / 3 Hence, ∆TF (BaCl2 ) = 3 x 2 / 2 = 3 TF = – 3◦C
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