Question

What is the change in the freezing point of water when 35.0 g of sucrose is dissolved in 300.0 g of water?

Kf of water = -1.86 degC/mol
molar mass sucrose = 342.30 g/mol
i value of sugar = 1

Answer 1

`DeltaT_f = -0.634` `""^"o""C"`

Explanation:

We're asked to find the freezing point depression of a solution.

To do this, we use the equation

`DeltaT_f = i·m·K_f`

where

• `DeltaT_f` is the change in freezing point temperature (what we're trying to find)

• `i` is the van't Hoff factor, which is given as `1` (and usually is `1` in the case of nonelectrolytes)

• `m` is the molality of the solution, which is

`"molality" = "mol solute"/"kg solvent"`

Convert the given mass of sucrose to moles using its molar mass:

`35.0cancel("g sucrose")((1color(white)(l)"mol sucrose")/(342.30cancel("g sucrose"))) = color(red)(0.102` `color(red)("mol sucrose"`

The molality is thus

`"molality" = color(red)(0.120color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg water") = color(green)(0.341m`

• `K_f` is the molal freezing point constant for the solvent (water), which is given as `-1.86` `""^"o""C/"m`

Plugging in known values, we have

`DeltaT_f = (1)(color(green)(0.341)cancel(color(green)(m)))(-1.86color(white)(l)""^"o""C/"cancel(m))`

`= color(blue)(ul(-0.634color(white)(l)""^"o""C"`

This represents by how much the freezing point decreases. The new freezing point of the solution is found by adding this value from the normal freezing point of the solvent (`0.0` `""^"o""C"` for water):

`"new f.p." = 0.0` `""^"o""C"` `- color(blue)(0.634` `color(blue)(""^"o""C"` `= ul(-0.634color(white)(l)""^"o""C"`