Suppose a student creates a solution by dissolving 0.5079 g of a non-dissociating, non-volatile unknown solute in 15.95 g cyclohexane. The freezing point change is −2.7°C. (a) Calculate the number of moles of solute in the solution. mol (b) Calculate the molar mass of the solute. g/mol Which one of the p-dihalobenzenes could this unknown be (p-difluoro-, p-dichloro, p-dibromo-, or p-diiodobenzene)?

a. 2.07×10⁻³ moles of solute

b. 244.9 g/mol

c. p-dibromobenzene, it is the nearest molar mass

Explanation:

Let's apply the colligative property of freezing point depression:

ΔT = Kf . m

where ΔT means Freezing T° of pure solvent - Freezing T° of solution

Kf is the cryoscopic constant and m, molality (moles of solute in 1kg of solvent)

Kf for the cyclohexane is 20.8 °C/m so, let's replace in the formula

2.7°C = 20.8°C/ m . m

2.7°C / 20.8 m/°C = 0.130 m

These are the moles in 1kg of solvent, but the mass of the solvent, is 15.95 g.

m = mol/kg → 0.130 m/kg . 0.01595kg = 2.07×10⁻³ moles

As this moles corresponds to 0.5079g, molar mass will be:

g/m → 0.5079 g / 2.07×10⁻³ moles = 244.9 g/mol

1.solid

2.carbon dioxide

3.helium

4d^55s^1 ( third choice)