At a certain temperature the vapor pressure of pure benzene (C6H6) is 0.930 atm. A solution was prepared by?

dissolving 10.0 g of a nondissociating, nonvolatile solute in 78.11 g of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm.

0.900 = 0.930 X
X =0.968 = moles benzene / moles benzene + moles solute
moles benzene = 78.11 g/78.11 g/mol = 1.00
0.968 = 1 / 1 + x
0.968 + 0.968 x = 1
0.968 x = 0.032
x = 0.0331 = moles solute
MM = 10.0 g/ 0.0331=302 g/mol