Question

Arrange the following solutions in order of decreasing freezing point: 0.20 m BaF2, 0.15 m C6H12, 0.10 m CrBr3, 0.15 m CH3CH2CH2COOH, 0.35 m KBr. (Enter just the formulas in your answer. Use the appropriate <, =, or > symbol to separate substances in the list.)

Answer 1

0.35 m
`KBr` > 0.20 m
`BaF_(2)` > 0.10 m
`CrBr_(3)` > 0.15 m
`CH_(3)CH_(2)CH_(2)COOH` > 0.15 m
`C_(6)H_(6)`

Step-by-step explanation:

For the given aqueous solutions:

`BaF_(2)` dissociates to form
`Ba^(2+)` and
`2F^(-)`

For 0.20 m of
`BaF_(2)`, the total concentration in ions = 3*0.20 = 0.60 m

`C_(6)H_(6)` does not dissociate. Therefore, the total concentration = 0.15 m.

`CrBr_(3)` dissociates to form
`Cr^(3+)` and
`3Br^(-)`

For 0.10 m of
`CrBr_(3)`, the total concentration in ions = 4*0.10 = 0.40 m

`CH_(3)CH_(2)CH_(2)COOH` dissociates to form
`CH_(3)CH_(2)CH_(2)COO^(-)` and
`H^(+)`.

For 0.15 m of
`CH_(3)CH_(2)CH_(2)COOH`, the total concentrations of ions = 2*0.15 = 0.30 m

`KBr` dissociates to for
`K^(+)` and
`Br^(-)`.

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
`KBr` > 0.20 m
`BaF_(2)` > 0.10 m
`CrBr_(3)` > 0.15 m
`CH_(3)CH_(2)CH_(2)COOH` > 0.15 m
`C_(6)H_(6)`