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Equimolal solutions of NaCl and BaCl_(2) are prepared in water. Freezing point of NaCl is foundto be -2^(@)C. What freezing point do you expect for BaCl_(2) solution? |
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Answer» Solution :`"i for NaCl = 2, i for BaCl"_(2)=3`. HENCE, `((DeltaT_(F))_("NACl"))/((DeltaT_(f))_("BaCl"_(2)))=(2)/(3)or(DeltaT_(f))_("BaCl"_(2))=(3)/(2)xx2=3^(@)" so that T"_(f)" for BACl"_(2)=-3^(@)C`. |
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