Respuesta :
Colligative property is a phenomenon where some physical properties of solvent (chemically pure substances i.e. cyclohexane) become disturbed as a result of adding other substances (usually a solute in this case).
Some of the colligative properties include:
- increasing(elevation) of boiling point
- decreasing(depression) of freezing point
The depression of a freezing point can be computed by using the formula:
[tex]\mathbf{\Delta T = K_fm}[/tex]
[tex]\mathbf{T_2-T_1= K_fm}[/tex]
where;
- T₁ = freezing point for a pure substance
- T₂ = substance freezing point in a non-volatile condition
- [tex]K_f[/tex] = freezing point constant
- m = molality of the solution
At standard condition for cyclohexane:
- Freezing point constant = 20.1° C.Kg/mol
- Pure cyclohexane freezing point = 6.5° C
Given that after adding the non-volatile unknown substance:
- Cyclohexane freezing point is 3.1° C
∴
Using the formula for the depression of freezing point, the molality of the unknown substance can be estimated as:
[tex]\mathbf{T_2-T_1= K_fm}[/tex]
[tex]\mathbf{(6.5-3.1)^0 C=(20.1 ^0 C \ kg/mol \times m)}[/tex]
[tex]\mathbf{(3.4)^0 C=(20.1 ^0 C \ kg/mol \times m)}[/tex]
[tex]\mathbf{m = \dfrac{3.4^0}{20.1 \ ^0 \ C \ kg/mol}}[/tex]
molality (m) = 0.17 mol/kg
Using the relation of molality = [tex]\mathbf{\dfrac{mass \ of \ the \ solute(unknown)}{ mass \ of \ the \ solvent(cyclohexane)\times molar mass(M)} }[/tex]
Making molar mass the subject of the formula and substituting the values, we have:
[tex]\mathbf{molar \ mass = \dfrac{0.750 \ g }{0.17 \ mol/kg\times 30.0 g}}[/tex]
[tex]\mathbf{molar \ mass = \dfrac{0.750\times 1000 kg }{0.17 \ mol/kg \times 30.0 g}}[/tex]
The molar mass of the unknown substance = 147.06 g/mol
Therefore, we can conclude that the molar mass of the unknown substance is 147.06 g/mol
Learn more about cyclohexane here:
https://brainly.com/question/5382898?referrer=searchResults