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)`.

Answer

`a. 1.5 mol kg^(-1)` `b. 1.448M` `c. 0.0263`

Explanation

`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`