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

1 Answer
Jul 31, 2017

Answer:

#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"#