Answer
$130\frac{g}{mol}$
Work Step by Step
We know that
$m=\frac{\Delta T_b}{K_b}$
We plug in the known values to obtain:
$m=\frac{0.55}{1.71}=0.32\frac{mol}{Kg}$
We can determine the number of moles of hydrocarbons as follows:
$0.095Kg\space solvent \times \frac{0.32mol\space hydrocarbon}{1Kg\space solvent}=0.030mol\space hydrocarbon$
Now $molar\space mass=\frac{3.75}{0.030}=130\frac{g}{mol}$
