Answer:
0.35 m [tex]KBr[/tex] > 0.20 m [tex]BaF_{2}[/tex] > 0.10 m [tex]CrBr_{3}[/tex] > 0.15 m [tex]CH_{3}CH_{2}CH_{2}COOH[/tex] > 0.15 m [tex]C_{6}H_{6}[/tex]
Explanation:
For the given aqueous solutions:
[tex]BaF_{2}[/tex] dissociates to form [tex]Ba^{2+}[/tex] and [tex]2F^{-}[/tex]
For 0.20 m of [tex]BaF_{2}[/tex], the total concentration in ions = 3*0.20 = 0.60 m
[tex]C_{6}H_{6}[/tex] does not dissociate. Therefore, the total concentration = 0.15 m.
[tex]CrBr_{3}[/tex] dissociates to form [tex]Cr^{3+}[/tex] and [tex]3Br^{-}[/tex]
For 0.10 m of [tex]CrBr_{3}[/tex], the total concentration in ions = 4*0.10 = 0.40 m
[tex]CH_{3}CH_{2}CH_{2}COOH[/tex] dissociates to form [tex]CH_{3}CH_{2}CH_{2}COO^{-}[/tex] and [tex]H^{+}[/tex].
For 0.15 m of [tex]CH_{3}CH_{2}CH_{2}COOH[/tex], the total concentrations of ions = 2*0.15 = 0.30 m
[tex]KBr[/tex] dissociates to for [tex]K^{+}[/tex] and [tex]Br^{-}[/tex].
Thus, for 0.35 m of KBr, the total concentration of ions = 2*0.35 m = 0.70 m
Therefore, arranging the aqueous solutions in order of decreasing freezing point:
0.35 m [tex]KBr[/tex] > 0.20 m [tex]BaF_{2}[/tex] > 0.10 m [tex]CrBr_{3}[/tex] > 0.15 m [tex]CH_{3}CH_{2}CH_{2}COOH[/tex] > 0.15 m [tex]C_{6}H_{6}[/tex]
