A solution of 0.85 g of an organic compound in 100.0 g of benzene has a freezing point of 5.16°C. What are the molality of the solution and the molar mass of the solute?

The molaility of the solution is 0.0625 m and the molar mass of the solute is 136 g/mol.

We have the following information from the question;

Mass of benzene = 100.0 g  or 0.1 Kg

Freezing point of solution =  5.16°C

Mass of solute = 0.85 g

We know that;

Freezing constant of benzene = 5.12 oC/molal

Freezing point of pure benzene = 5.48 oC

From;

ΔT = K m i

ΔT = Freezing point depression

K = Freezing constant

m = molality

i = Van't Hoff factor

ΔT = 5.48 oC - 5.16°C = 0.32°C

0.32°C = 5.12 oC/m × m × 1

m = 0.32°C/5.12 oC/m × 1

m = 0.0625 m

Also;

molality = Number of moles/Mass of solution in kilograms

Number of moles = molality × Mass of solution in kilograms

Number of moles =  0.0625 m ×  0.1 Kg

Number of moles = 0.00625 moles

Again;

Number of moles = mass/molar mass

Molar mass = mass/Number of moles

Molar mass =  0.85 g/ 0.00625 moles

Molar mass = 136 g/mol

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The molar mass of the solute(organic compound)=128.012 g/mol

Solution properties are the properties of a solution that don't depend on the type of solute but only on the concentration of the solute.  

Solutions from volatile substances have a higher boiling point and lower freezing points than the solvent  

For freezing point can be formulated

K = molal freezing point constant  

m = molal solution  

ΔT for solution (freezing point for Benzene = 5.5 °C) :

molal  : the number of moles of solute in 1 kg of solvent  

mass of organic compound = 0.85 g

it's filtration, but this is an assumption so my apologies if i got this