Question

Which solution will exhibit the GREATEST freezing point depression for `1*mol*L^-1` aqueous solutions of: `"(i) propanol; (ii) potassium bromide; (iii) calcium bromide?"`

Answer 1

`"Freezing point depression"` is a so-called `"colligative property"`.......and I think you mean to consider `CaBr_2` NOT `CrBr_2`.

Explanation:

And a colligative property depends on the NUMBER of solute particles, not their identity. Solute-solvent interactions tend to (i) INCREASE the boiling point of solutions by an amount proportional to the number (i.e. concentration) of solute particles, and (ii) depress the melting point of solutions RELATIVE to pure solvent, again by reason of solute-solvent interactions.

In each solution, the concentration is `1*mol*L^-1` with respect to the solute. However, some of the solutes are electrolytes:

`"propanol"(l) stackrel(H_2O)rarr"propanol"(aq)`

`KBr(s) stackrel(H_2O)rarrK^(+) + Br^(-)`

`CaBr_2(s) stackrel(H_2O)rarrCa^(2+) + 2Br^(-)`

All the solutions are `1*mol*L^-1` with respect to solute. However, for the calcium bromide solution (if that is what you consider!), there are THREE particles in solution rather than TWO (for `KBr(aq)`) or ONE for propanol.

So in terms of melting point, the calcium bromide solution should be LOWEST, and next is the potassium bromide solution, and the next is the propanol solution. The elevation of boiling point follows the reverse order.