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
[tex]\tt \Delta T_f=K_f.m[/tex]
K = molal freezing point constant
m = molal solution
ΔT for solution (freezing point for Benzene = 5.5 °C) :
[tex]\tt \Delta T_f=5.5-5.16=0.34[/tex]
[tex]\tt m=\dfrac{\Delta T_f}{K_f}\rightarrow K_f~for~Benzene=5.12^oC/m\\\\m=\dfrac{0.34}{5.12}=0.0664[/tex]
molal : the number of moles of solute in 1 kg of solvent
[tex]\tt 0.0664=\dfrac{mol}{0.1~kg~benzene}\\\\mol=0.0664\times 0.1=0.00664[/tex]
mass of organic compound = 0.85 g
[tex]\tt MW=\dfrac{mass}{mol}=\dfrac{0.85}{0.00664}=128.012~g/mol[/tex]
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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