An organic solute was dissolved in napthalene such that the freezing point dropped by 7.9 degrees. What is the molality of the solution?
1 Answer
#"1.13 mol/kg"# to three sig figs (although you only have two...).
Well, you either haven't given us the freezing point depression constant, or you weren't given it. I also assume your degrees are
Either way, we could either look that up, or use the enthalpy of fusion from NIST (Sharma, Gupta, et al., 2008) and the following expressions.
From Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-3, the freezing point depression for non-electrolytes is given by:
#\mathbf(DeltaT_"f" = T_"f" - T_"f"^"*" = -K_fm)#
#\mathbf(K_f = (M_"solvent")/("1000 g/kg")(R(T_"f"^"*")^2)/(DeltabarH_"f"))# where:
#T_"f"^"*" = "353.41 K"# is the freezing point in#"K"# of the pure solution (no solute)#T_"f"# is the freezing point in#"K"# of the solution (with solute)#M_"solvent"# is the molar mass of the solvent in#"g/mol"# .#R# is the universal gas constant:#"8.314472 J/mol"cdot"K"# #DeltabarH_"f" ~~ "19.1 kJ/mol"# is the molar enthalpy of fusion, in#"kJ/mol"# (hence, "molar").#K_f > 0# is known as the freezing point depression constant in#"K"cdot"kg/mol"# .#m# is the molality, which is#"mols solute"/"kg solvent"# .
So, we could obtain
#K_f = (128.1705 cancel"g""/mol")/(1000 cancel"g""/kg")((0.008314472 cancel"kJ""/"cancel"mol"cdotcancel"K")("353.41 K")^cancel(2))/(19.11 cancel"kJ""/"cancel"mol")#
#= "6.97 K"cdot"kg/mol"#
while the first reference gives
#-"7.9 K" = DeltaT_f = T_f - T_f^"*" = -K_fm#
This gives a molality of:
#color(blue)(m) = (DeltaT_f)/(-K_f) = (-7.9 cancel"K")/(-6.97 cancel"K"cdot"kg/mol")#
#=# #color(blue)("1.13 mol/kg")#
although we only have two sig figs, apparently...
