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...