An aqueous solution has a boiling point of 102.45°C. What is its freezing point? ( Kf = 1.86°C m-1 and Kb = 0.51°C m-1 )

The freezing point for the solution is -8.9°C

Explanation:

We must combine the two colligative properties of freezing point depression and boiling point elevation to solve this excersise.

ΔT = Kb . m

ΔT = T° boiling point of solution - T° boiling of pure solvent

Kb = Ebullioscopic constant

m = molality

102.45°C - 100°C = 0.51°C/m . m

2.45°C / 0.51 m /°C = m → 4.80 mol/kg

We found out the molality in the boiling point elevation to replace at the freezing point depression formula

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryoscopic constant

0°C - T° freezing solution = 1.86°C/m . 4.80 m

T° freezing solution = - 8.9°C

Answer is: carbon atoms can be arranged straight chain, branched chain and ring.
Hydrocarbon is an organic compound consisting of hydrogen and carbon.
In straight chain carbon atoms are bonded with only two other carbon atoms in elongate chain.
In branched chain at least one carbon is bonded with more than two other carbon atoms.
Ring chain hydrocarbons are cyclic hydrocarbons.