A solution of ethanol in water boils at 101.5 degrees C. what it tne molarity? Can you show me the setup i would use??

I did the first time you posted this problem.
delta T = i*Kb*m
delta T = 101.5 - 100 = 1.5
i = 1
kb = 0.51
m = ??

m equals molarity
i thought it equals moles!!!!!

m = molality.
the abbrev for moles is mols.

then for those other problems we converted molarity into grams??
and the final answer for the ethanol one is 2.94

2.94 m = 2.94 molal is what I obtained. Yes, for the other problems we converted molality into grams from molality = mols solute/kg solvent.

WHAT ARE THE THREE ANSWERS YOU GOT!!
C6H1206= 84.4
CACL2= 13.92

I'm sorry for the confusion. Let me clarify. In this specific problem, we are trying to find the molarity of the ethanol solution. Molarity is defined as the number of moles of solute per liter of solution.

To find the molarity, we need to use the equation:

delta T = i * Kb * m

Where:
delta T = boiling point elevation (difference between the boiling point of the solution and the boiling point of the pure solvent)
i = van't Hoff factor (the number of particles the solute dissociates into in solution)
Kb = molal boiling point constant for the solvent
m = molality of the solution (moles of solute per kilograms of solvent)

In this case, we know that:
delta T = 101.5 - 100 = 1.5 degrees Celsius (since the boiling point of water is 100 degrees Celsius)
i = 1 (since ethanol does not dissociate in water)
Kb = 0.51 (molal boiling point constant for water)
m = ?

Now, to calculate m, we rearrange the equation and solve for m:

m = delta T / (i * Kb)

Substituting the given values:

m = 1.5 / (1 * 0.51)
m ≈ 2.94 mol/kg

So, the molality of the ethanol solution is approximately 2.94 mol/kg, not the molarity.

In previous discussions, we might have converted molality into grams using the formula molality = moles solute / kg solvent, but in this particular problem, we are finding molality, not molarity.

I apologize for any confusion caused. Let me know if you have any further questions.