Question #a4a55

To solve this problem involving freezing-point depression, we can use the equation

#DeltaT_f = m • i • K_f#

Where

• #DeltaT_f# is the change in freezing point temperature (a positive quantity, although the temperature decreases)

• #m# is the molality of the solution, equal to the moles of solute per kilogram of solvent (we'll find this)

• #i# is the van't Hoff factor of the solute, which is given as #1# (it is this value for basically all nonelectrolytes)

• #K_f# is the molal boiling point constant of the solvent, which is given for water as #1.86# #m//""^"o""C"#

We have to find the molality of the sucrose solution.

The number of moles can be found using the given molar mass (#342.30# #"g/mol"#) and mass (#35.0# #"g"#):

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

And there are

#300.0cancel("g H"_2"O")((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = 0.3000# #"kg H"_2"O"#

So the molality is

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

#= (0.102color(white)(l)"mol sucrose")/(0.3000color(white)(l)"kg H"_2"O") = color(red)(0.341m#

Now, plugging in our known values, we have

#DeltaT_f = (0.341cancel(m))(1)(1.86(""^"o""C")/(cancel(m))) = color(blue)(0.634^"o""C"#

So option (c) is the correct answer. I was taught that the #DeltaT_f# quantity is a positive value (although I disagree with this sometimes!), but just know that this is how much the freezing point lowers.

And since the freezing point lowers, this value is what you would subtract from the normal freezing point of water (#0^"o""C"#):

#0^"o""C" + (-0.634^"o""C") = color(blue)(-0.634^"o""C"#

And the new freezing point of the solution is thus #color(blue)(-0.634^"o""C"#.