InterviewSolution
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InterviewSolution
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In a `0.2` molal aqueous solution of weak acid `HX` (the degree of dissociation `0.3`) the freezing point is (given `K_(f) = 1.85 K molality^(-1)`):A. `+0.480^(@)C`B. `-0.480^(@)C`C. `-0.360^(@)`D. `+0.360^(@)` |
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Answer» Correct Answer - 2 We have `(DeltaT_(f))_("observed")=(K_(f))_("observed")` molality `" "HX(aq)hArrH^(+)(aq)+X^(-)(aq)` `{:("Moles before ionization",+,1 mol,0 mol,0 mol),("Moles after ionization",-,1-alpha mol,alpha mol,alpha mol):}` `i=("Total moles after ionization")/("Total moles before ionozation")` `=(1-alpha+alpha+alpha)/1=1+alpha` `=1+0.3` `=1.3` we also have `i=(K_(f) ("observed"))/(K_(f)("calculated"))` or `K_(f) ("observed")=i K_(f)("calculate")` `=(1.3)(1.85)` `=2.405` Thus `DeltaT_(f)=(2.405 K kg mol^(-1))(0.28 mol kg^(-1))` `=0.481 K` or `0.481^(@)C` Hence Freezing point of solution =Freezing point of water `-DeltaT_(f)` `=(0.000^(@)C)- (0.481^(@)C)` `= -0.481^(@)C` |
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