The freezing point of the solution will be equal to #0.53^@"C"#.
SIDE NOTE The cryoscopic constant, #K_f#, for benzene is actually equal to #5.12^@"C/molal"#, not #4.90^@"C/molal"#, so I solved the problem using the correct value.
If you want, you can redo the calculations with the value you have/were given.
So, you know that you're dealing with a solution containing naphthalene, your solute, and benzene, your solvent. The first thing you need to do is determine the mass of benzene by using its volume and density
#rho = m/V => m = rho * V#
#m_"benzene" = 0.877"g"/cancel("mL") * 722cancel("mL") = "633.2 g"#
The equation for freezing-point depression looks like this
#DeltaT_"f" = i * K_f * b#, where
#K_f# - the cryoscopic constant - depends on the solvent;
#b# - the molality of the solution;
#i# - the van't Hoff factor - the number of ions per individual molecule of solute.
#DeltaT_"f"# - the freezing point depression - is defined as
#T_"f"^0 - T_"f sol"#.
In your case, the van't Hoff factor will be equal to 1 because you'rea dealing with molecular compounds which do not dissociate.
Use naphthalene's molar mass to determine how many moles you're adding to the solution
#78.8cancel("g") * "1 mole"/(128.16cancel("g")) = "0.6149 moles"#
The solution's molality will thus be
#b = n_"naphthalene"/m_"benzene"#
#b = "0.6149 moles"/(633.2 * 10^(-3)"kg") = "0.971 molal"#
Now you have all you need to solve for #DeltaT_"f"#
#DeltaT_"F" = 1 * 5.12^@"C"/cancel("molal") * 0.971cancel("molal") = 4.97^@"C"#
This means that the freezing point of the solution is
#DeltaT_"f" = T_"f"^0 - T_"f sol" => T_"f sol" = T_"f"^0 - DeltaT_"f"#
#T_"f sol" = 5.50 - 4.97 = color(green)(0.53^@"C")#
