# If #"0.64 g"# of napthalene (#M = "128.1705 g/mol"#) is dissolved in #"100 g"# a certain solvent (#T_f^"*" = 6^@ "C"#), whose #K_f = 20^@ "C"cdot"kg/mol"#, what is the new freezing point of the solvent?

##
#a)# #4^@ "C"#

#b)# #5^@ "C"#

#c)# #-1^@ "C"#

#d)# #7^@ "C"#

##### 1 Answer

That's a high

The **change in freezing point** is given by:

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

#T_f# is thefreezing pointof the solution.#"*"# indicates pure solvent.#i# is thevan't Hoff factorof the solute.#i >= 1# , and corresponds to the effective number of particles per formula unit of solute.#K_f# is thefreezing point depression constantof the solvent.#m# is themolalityof the solution in#"mol solute/kg solvent"# .

#m = (0.64 cancel"g Napthalene" xx "1 mol"/(128.1705 cancel"g Napthalene"))/(100 cancel"g solvent" xx "1 kg"/(1000 cancel"g solvent")#

#=# #"0.050 mol solute/kg solvent"#

Therefore, the **change** in freezing point is:

#DeltaT_f = -(1)(20^@ "C"cdot"kg/mol")("0.0499 mol/kg")#

#= -0.9987^@ "C" -> -1^@ "C"#

Thus, the **new freezing point** is:

#color(blue)(T_f) = DeltaT_f + T_f^"*" = (-0.9987^@ "C") + (6^@ "C")#

#~~ color(blue)(5^@ "C")#