What is the freezing point of an aqueous solution that boils at 105.4 ∘c?

Answer : -19.6

Explanation : We can calculate this using the combine equation of boiling point and depression in freezing point and obtain a combined equation which is

∆T(f) = ∆T(b) x {K(f) / K(b)} ;

Where ∆T(f) is depression in freezing point,

∆T(b) is elevation in boiling point,

K(f) is cryoscopic constant for water it is 1.86 

K(b) is ebullioscopic constant for water it is 0.512

So we can solve the equation,

∆T(f) = (105.4 -100) X = 19.6 

So when we have the freezing point depression as 19.6,

we know water freezes at 0° C so the freezing point will be 0 - 19.6 = -19.6 °C

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.