# 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.)

##### 1 Answer

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

Given the freezing point depression equation:

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

Since we don't have *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,

#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

#"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.