1,15g of a non-volatile compound is dissolved in 75,0g of benzene. The boiling point of the solution 0,125ºC has a value greater than the corresponding pure benzene. Calculate the molecular weight of the compound.

I think you need to re-word the question. I find it difficult to believe that a solution boiling at 0.125C is greater than that of pure benzene. Do you know the b.p. of pure benzene. I think it's about 80 C.

To calculate the molecular weight of the compound, we need to use the formula for boiling-point elevation.

The boiling point elevation (∆Tb) can be calculated using the equation:

∆Tb = Kb * m

where Kb is the molal boiling point elevation constant (for benzene, Kb = 2.53 °C/m) and m is the molality of the solution.

Molality (m) is defined as the moles of solute per kilogram of solvent. To calculate molality, we first need to calculate the number of moles of the compound and the number of moles of benzene.

To find the number of moles of the compound, we can use the given mass and the compound's molar mass (M):

moles of compound = mass of compound / molar mass

Next, we calculate the moles of benzene using its molar mass and the given mass:

moles of benzene = mass of benzene / molar mass of benzene

Once we have the number of moles of the compound and benzene, we can calculate molality (m):

molality = moles of compound / mass of benzene (in kg)

Finally, we can then substitute molality (m) and boiling-point elevation constant (Kb) into the boiling point elevation equation to find ∆Tb:

0.125 °C = Kb * molality

Now, we can solve for molality:

molality = 0.125 °C / Kb

Once we have found molality, we can determine moles of solute per kilogram of solvent. The solute in this case is the compound, and the solvent is benzene.

Finally, to find the molecular weight of the compound, divide the mass of the compound by the calculated number of moles.

molecular weight = mass of compound / moles of compound

By following this procedure, you will be able to calculate the molecular weight of the compound.