A solution was prepared by dissolving 0.42 mol hexane into 500g CCl4. What is the change in freezing point of this solution? The Kf for CCl4 is 30.0oC/m. The freezing point of CCl4 is -22.92 °C.

Freezing T° of solution = - 48.12°C

Explanation:

The colligative which has to be used for this case is the freezing point depresison ( ΔT = Kf . m )

ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Crysocopic constant

m = molality (mol/kg)

We determine the molality (moles of solute in 1kg of solvent)

We convert the mass of CCl₄ from g to kg → 500 g . 1kg / 1000g = 0.5kg

0.42 mol of hexane / 0.5 kg of CCl₄ = 0.84 mol/kg

Let's replace data: -22.92°C - Freezing T° of solution = 30°C/m . 0.84 m

Freezing T° of solution = - ( 30°C/m . 0.84 m + 22.92°C) → - 48.12°C

the increase of protons   hope this

its first and third is your answer i did the qustion the fourth isnt right