Answer
See the answer below.
Work Step by Step
Freezing:
$\Delta T_f=K_fm_f\rightarrow m_f=(3.1-5.50)^{\circ}C\div-5.12^{\circ}C/m=0.469\ m$
Boiling:
$\Delta T_b=K_bm_b\rightarrow m_b=(82.6-80.10)^{\circ}C\div2.53^{\circ}C/m=0.988\ m$
Since the ratio of the two is approximately two, it's likely that at lower temperatures the prevalent form of benzoic acid is in dimers, caused by the hydrogen bonding (similar to what happens with acetic acid).
