## Chemistry: The Molecular Science (5th Edition)

$52.5\,g/mol$
$T_{f}(solution)= 5.15^{\circ} C$ (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=$\frac{mass\,of\, solute}{number\, of\, moles}=\frac{0.180\,g}{0.00343\,mol}=52.5\,g/mol$