Question

Question #09585

Answer 1

The solution in which you dissolve the sodium chloride will have a lower freezing point.

This happens because freezing-point depression is a colligative property that depends on the concentration of particles in solution, not on what those particles are.

Mathematically, this is expressed as

`DeltaT_f = K_f * b_F * i`, where

`DeltaT_f` - the freezing-point depression, defined as the freezing temperature of the pure solvent minus the freezing point of the solution;
`K_f` - the cryoscopic constant, which depends solely on the solvent;
`b_F` - the molality of the solution;
`i` - the van't Hoff factor, which takes into account the number of particles produced by a compound when dissolved in solution.

The van't Hoff factor is the key to why the sodium chloride solution will have a lower freezing point.

Sodium chloride is a strong electrolyte, which means it dissociates completely in aqueous solution to give sodium cations and chloride anions

`NaCl_((aq)) -> Na_((Aq))^(+) + Cl_((aq))^(-)`

Notice that 1 mole of `NaCl` produces 1 mole of `Na^(+)` and 1 mole of `Cl^(-)`. This means that, regardless of how many moles of `NaCl` you add to the solution, you'll have twice as many moles of ions present.

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As a result, the van't Hoff factor for `NaCl` will be 2.

This is not the case for ethanol. When placed in aqueous solution, a non-electrolyte does not dissociate to form ions. As a result, its van't Hoff factor will be equal to 1.

To compare the freezing points of the two solutions, just use

`DeltaT_"f NaCl" = K_f * 0.5 * 2 = K_f`

`DeltaT_"f ethanol" = K_f * 0.75 * 1 = 0.75 * K_f`

or

`T_"f pure water" - T_"f NaCl" = K_f => T_"NaCl" = T_"f pure water" - K_f`

`T_"f ethanol" = T_"f pure water" - 0.75 * K_f`

As you can see,

`T_"f NaCl" < T_"f ethanol"`