What mass of cholesterol should be dissolved in #"295 g"# of benzene to lower the freezing point by #"0.450 K"#? (#M = "386.6 g/mol"# for cholesterol and #"78.11 g/mol"# for benzene.)

I got

#10.0_4# #"g"#

of cholesterol to three sig figs (the subscript indicates the first digit of which we are not certain) to reduce the normal freezing point of benzene to #5.05^@ "C"#.

Given the freezing point depression equation:

#bb(DeltaT_f = T_f - T_f^"*" = -iK_fm)#,

where:

• #T_f# is the freezing point, and #"*"# indicates pure solvent.
• #i# is the van't Hoff factor, and is #1# for nonelectrolytes.
• #K_f# (not given) is the freezing point depression constant of benzene, the solvent.
• #m# is the molality of the solution, #"mol solute/kg solvent"#.

Since we don't have #K_f#, we'll have to get it from (Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3):

#\mathbf(K_f = (M_"solvent")/("1000 g/kg")(R(T_f^"*")^2)/(DeltabarH_f))#

where #M# is the molar mass in #"g/mol"#, #R# is the universal gas constant, and #DeltabarH_f# is the molar enthalpy of fusion in #"J/mol"#.

From NIST, #DeltabarH_f = "9.87 kJ/mol"#, so the #K_f# is:

#K_f = ("78.11 g/mol")/("1000 g/kg") ("8.314472 J/mol"cdot"K" cdot (5.50 + 273.1"5 K")^2)/("9870 J/mol")#

#= "5.11 K"cdot"kg/mol"#

This could be verified here... This allows us to get the molality:

#color(green)(m) = (DeltaT_f)/(-iK_f)#

#= (-"0.450 K")/(-(1)("5.11 K"cdot"kg/mol"))#

#=# #color(green)"0.0881 mol solute/kg solvent"#

Since we know that #"295 g benzene"#, or #"0.295 kg"#, were used, and we know that cholesterol has a molar mass of #"386.6 g/mol"#:

#"0.0881 mol solute"/cancel"kg solvent" xx 0.295 cancel"kg solvent"#

#=# #"0.0260 mol cholesterol"#

As a result, we have this mass of solute dissolved in benzene:

#0.0260 cancel"mols cholesterol" xx "386.6 g"/cancel"mols cholesterol"#

#=# #color(blue)("10.0 g cholesterol")# to three sig figs.