Two solutions are prepared using the same solute: Solution A: 0.27 g of the solu

Question

Two solutions are prepared using the same solute:
Solution A: 0.27 g of the solute dissolves in 27.4 g of t-butanol.
Solution B: 0.23 g of the solute dissolves in 24.8 g of cyclohexane.
Which solution has the greatest freezing point change? Show calculations and explain.

The kf (freezing point constant) of t-butanol = 9.1
The kf (freezing point constant) of cyclohexane= 20.0

The formula I am using is Delta Tf (or Change in Freezing Point)= kf (freezing point constant) x m (molality of the solute in solution)

The formula for molality= mol solute / kg solvent or (mass/molar mass)/kg solvent

This is how I solved for the freezing point change of Solution A:
M will stand for molar mass in my equation.

Delta Tf= (9.1) x (0.27 g/ M)/(27.4 x 10-3 kg)

I used this equation because I do not know how many moles of the soluteare present in the solution. I just know the mass for the solute.
Delta Tf= 89.67/M

This is how I solved for the freezing point change of Solution B:

Delta Tf= (20.0) x (0.23 g/M)/24.8 x 10-3 kg)
Detla Tf= 185.5/M

I really have no idea if this is even the right way to solve this kind of problem. The other questions posted for this type of problem were unclear. Please leave a detailed explanation,if possible.

Explanation / Answer

yes you have done it absolutely right.. since m will have a fixed value the depression in b case will be much larger because it has a gretaer denominator this is the explanation plz rate me first

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