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Calculate the `(a)` molality, `(b)` molartiy, and `(c)` mole fraction of `KI` if the density of `20% (` mass `//` mass `)` aqueous `KI` is `1.202 g m L^(-1)`. |
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Answer» Correct Answer - `a. 1.5 mol kg^(-1)` `b. 1.448M` `c. 0.0263` `W_(2)=20g,W_(sol)=100g` `W_(1)=100-20=80g` `m=(W_(2)xx1000)/(Mw_(2)xxW_(1))=(20xx1000)/(166xx80)=1.5mol Kg^(-1)` `b.` `Mw_(2)` of `Kl=39+127=166g mol^(-1)` `M+(% by mass xx10xxd_(sol))/(Mw_(2))=(20xx10xx1.202)/(166)` `=1.448M` `c.` `CHMi_(2)=(n_(2))/(n_(1)+n_(2))=(W_(2)//Mw_(2))/(W_(1)//Mw_(1)+W_(2)//Mw_(2))` `(20//166)/(80//18+20//166)` `=(0.12)/(4.44+0.12)=(0.12)/(4.56)=0.0263` |
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