An organic solute was dissolved in napthalene such that the freezing point dropped by 7.9 degrees. What is the molality of the solution?

#"1.13 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 #""^@ "C"#...

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# ourselves:

#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 #6.94#. The change in freezing point was given:

#-"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...