#### Answer

$52.5\,g/mol$

#### Work Step by Step

$T_{f}(solution)= 5.15^{\circ} C$ (given)
$T_{f}(solvent)=5.5^{\circ} C$ (freezing point of benzene)
$\Delta T_{f}=T_{f}(solution)-T_{f}(solvent)$
$=5.15^{\circ} C-5.5^{\circ} C= -0.35^{\circ} C$
$K_{f}$ for benzene=$-5.10^{\circ} C\,kg/mol$
But $\Delta T_{f}= K_{f}\times m_{solute}$
$\implies molality\,\,m_{solute}=\frac{\Delta T_{f}}{K_{f}}$
$=\frac{-0.35^{\circ} C}{-5.10^{\circ} C\,kg/mol}=0.0686\,mol/kg$
As molality=$\frac{moles\,of\,solute}{mass\,of\, solvent\,in\,kg}$, we have
Moles of solute= $m_{solute}\times$mass of solvent in kg
$=0.0686\,mol/kg\times0.050\,kg=0.00343\,mol$
Molar mass of solute=$\frac{mass\,of\, solute}{number\, of\, moles}=\frac{0.180\,g}{0.00343\,mol}=52.5\,g/mol$