Answer
$C_{14}H_{10}$
Work Step by Step
Boiling point elevation: $80.34-80.10=0.24^{\circ}C$
Molality: $\Delta T_b=K_bm\rightarrow m=0.24^{\circ}C\div 2.53^{\circ}C/m=0.095\ mol/kg$
Number of moles: $0.095\ mol/kg\times (30.0/1000)kg=2.85\times10^{-3}\ mol$
Molar mass: $0.500\ g\div 2.85\times10^{-3}\ mol=175.69\ g/mol$
Molar mass of empirical formula: $89.11\ g/mol$
Molar mass ratio: $175.69/89.11=1.97\approx 2$
Molecular formula: $2\times empirical\ formula=C_{14}H_{10}$
