Answer
$3.4\times10^{2}\,g/mol$
Work Step by Step
$\Delta T_{f}=1.3^{\circ}C$
$K_{f}=1.86\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
Molality $m=\frac{\Delta T_{f}}{K_{f}}=\frac{1.3}{1.86}\frac{mol\,solute}{kg\,solvent}=0.6989\frac{mol\,solute}{kg\,solvent}$
$m=\frac{\text{Moles solute}}{\text{Kilograms solvent}}\implies \text{Moles solute}=m\times\text{Kilograms solvent}$
$=(0.6989\times0.1500)mol=0.104835\,mol$
Molar mass= $\frac{\text{Mass in grams}}{\text{Number of moles}}=\frac{35.9\,g}{0.104835\,mol}=3.4\times10^{2}\,g/mol$
