# 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 thepure solution(no solute)#T_"f"# is the freezing point in#"K"# ofthe solution(with solute)#M_"solvent"# is themolar massof the solvent in#"g/mol"# .#R# is theuniversal gas constant:#"8.314472 J/mol"cdot"K"# #DeltabarH_"f" ~~ "19.1 kJ/mol"# is themolar enthalpy of fusion, in#"kJ/mol"# (hence, "molar").#K_f > 0# is known as thefreezing point depression constantin#"K"cdot"kg/mol"# .#m# is themolality, 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...