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An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:A. 0.0585 KB. 0.0512 KC. 1.864 KD. 0.0265 K |
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Answer» Correct Answer - B `(DeltaT_f)/(DeltaT_b)=K_f/K_b DeltaT_b=(K_bxxDeltaT_f)/K_f` `DeltaT_b=(0.512xx0.186)/1.86=0.0512` |
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