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
Kf of water = 1.86 degC/mol
molar mass sucrose = 342.30 g/mol
i value of sugar = 1
1 Answer
Answer:
Explanation:
We're asked to find the freezing point depression of a solution.
To do this, we use the equation
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
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 (