Answer:
The molar mass of the unknown compound is 202.40 g/mol.
Explanation:
Let the molar mass of compound be M.
The depression in freezing point is given by:
[tex]\Delta T_f=K_f\times m[/tex]
[tex]\Delta T_f=T-T_f[/tex]
where,
[tex]\Delta T_f[/tex]= change in boiling point = 0.81 K
[tex]T_f[/tex]=Boiling point of the solution = 3.77°C
T = freezing point of the pure solvent here benzene =5.48°C
[tex]K_f [/tex]= freezing point constant = 5.12°C/m
m = molality =[tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\text{Mass of solvent in kg}}[/tex]
[tex]\Delta T_f=5.48^oC-3.77^oC=1.71^oC[/tex]
[tex]1.71^oC=5.12^oC/m\times \frac{33.8 g}{M\times 0.500 kg}[/tex]
M = 202.40 g/mol
The molar mass of the unknown compound is 202.40 g/mol.
