The answer is #6.33# #"g"# of #MnI_2# must be used to get that drop in freezing point temperature.
The mathematical equation for boiling point depression, which is a colligative property, is this
#DeltaT_(f) = K_(f) * m * i#, where
#DeltaT_f# - the freezing point depression, defined as the difference between the freezing temperature of the pure solvent minus the freezing temperature of the solution, #T_f("pure [solvent](http://socratic.org/chemistry/solutions-and-their-behavior/solvent)") - T_f("solution")#;
#K_f# - the molal freezing point depression constant of the solvent -#1.86^@C*"kg/mol"# for water;
#m#- the molality of the solution;
#i# - the number of dissolved particles (Van't Hoff Factor).
You must use the equation above to determine the molality of the solution, i.e. the moles of #MnI_2# you have in that amount of water. The moles of #MnI_2# will then get you the mass you need to determine. So, you know that #DeltaT_(f)# is equal to #0.4500^@C#.
Since #MnI_2# is soluble in water, it will dissociate into #Mn^(2+)# and #2I^(-)# ions, which will give you a Van't Hoff factor of 1 + 2 = 3. Every molecule of #MnI_2# creates 3 ions when placed in water, 1 #Mn^(2+)# cation and 2 #I^(-)# anions. So,
#DeltaT_(f) = K_(f) * m * i => m = (DeltaT_(f))/(K_(f) * i)#
#m = (0.4500^@C)/(1.86^@C * (kg)/(mol) * 3) = 0.0806# #"molal"#
You know that molality is defined as moles of solute divided by mass of solution (in kg). This will get you
#m = n_("solute")/m_("solvent") = n_(MnI_2)/(m_("water")) => n_(MnI_2) = m * 254.0 * 10^(-3)kg#
#n_(MnI_2) = 0.0806 * 254.0 * 10^(-3)# #"kg" = 0.0205# #"moles"#
This means that the mass of #MnI_2# needed is
#m_(MnI_2) = n_(MnI_2) * "molar mass"#
#m_(MnI_2) = 0.0205# #"moles" * 308.7# #"g/mol" = 6.33# #"g"#
