Answer is: the molar mass of unknown molecular compound is 245.67 g/mol.
m(compound) = 5.65 g.
m(benzene) = 110 g ÷ 1000 g/kg = 0.11 kg.
ΔT = 5.45°C - 4.39°C.
ΔT = 1.06°C.
Kf(benzene) = 5.07°C/m.
M(compound) = Kf
· m(compound) / m(benzene) · ΔT.
M(compound)= 5.07°C/m · 5.65 g / (0.11 kg · 1.06°C).
M(compound) = 245.67 g/mol.
Answer:
238.91 g/mol is the molar mass of the solute.
Explanation:
Mass of solute = 5.65 g
Molar mas of solute = M
Mass of solvent = 110.0 g = 0.110 kg
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]\Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] =Depression in freezing point
[tex]K_f[/tex] = freezing point constant of solvent= 5.07°C/m
1 =van't Hoff factor
m = molality
Freezing point constant of benzene ,T= 5.45°C/m
T = 5.45°C ,[tex]T_f[/tex] =4.39 °C
[tex]\Delta T_f=T-T_f=5.45^oC-4.39 ^oC=1.09^oC[/tex]
[tex]1.09^oC=1\times 5.07^oC kg/mol\times \frac{5.65 g}{M\times 0.110 kg}[/tex]
[tex]M=1\times 5.07^oC kg/mol\times \frac{5.65 g}{1.09^oC\times 0.110 kg}[/tex]
i = 1
M = 238.91 g/mol
238.91 g/mol is the molar mass of the solute.
