Given that the vapor pressure of water is 17.54 torr at 20 C, calculate the vapor-pressure lowering of aqueous solutions that are 2.40 m in (a) sucrose, C12H22O11, and (b) aluminum chloride. Assume 100% dissociation for electrolytes.

Sucrose Delta P = __ torr
Aluminum Chloride Delta P = __ torr

here's what I did..
2.4/(2.4+55.56) = 0.041
p = (0.41)(17.54) = 0.726 torr sucrose?

p = (3)(0.726) = 2.178 torr for aluminum chloride?

Hmm, I can see that you're trying your best with those calculations, but let me just clarify something first. Are you sure you want me, Clown Bot, to calculate something? They don't call me "Clown Bot" for my mathematical prowess, you know!

To calculate the vapor-pressure lowering of aqueous solutions, you need to use the equation:

ΔP = Xsolute * Psolvent

where ΔP is the vapor-pressure lowering, Xsolute is the mole fraction of the solute, and Psolvent is the vapor pressure of the pure solvent.

For part (a), sucrose (C12H22O11), you correctly calculated the mole fraction of sucrose in the solution as 0.041 (2.4 / (2.4 + 55.56)). The vapor pressure of water at 20°C is given as 17.54 torr. So, the calculation would be:

ΔP = 0.041 * 17.54 = 0.71814 torr

Rounding to the correct number of significant figures, the vapor-pressure lowering of the sucrose solution is approximately 0.718 torr.

For part (b), aluminum chloride (AlCl3), you mentioned that it is assumed to be 100% dissociated in water. Therefore, you need to consider the number of ions that contribute to the vapor-pressure lowering. Aluminum chloride dissociates into three ions, one aluminum ion (Al3+) and three chloride ions (3Cl-). The final equation would be:

ΔP = 3 * XAlCl3 * Psolvent

Since it is not given, you may assume Psolvent as the vapor pressure of pure water at 20°C, which is 17.54 torr. If the mole fraction of aluminum chloride is 1, then:

ΔP = 3 * 1 * 17.54 = 52.62 torr

So, the vapor-pressure lowering of the aluminum chloride solution would be 52.62 torr.

Thus, the correct vapor-pressure lowerings are:
(a) Sucrose: ΔP = 0.718 torr
(b) Aluminum Chloride: ΔP = 52.62 torr

To calculate the vapor-pressure lowering of aqueous solutions, you can use the equation ΔP = X * P°, where ΔP represents the vapor-pressure lowering, X is the mole fraction of the solute, and P° is the vapor pressure of the solvent (water in this case).

(a) Sucrose (C12H22O11)
To calculate the mole fraction of sucrose, you correctly used the formula X = moles of solute / total moles of solution. However, there is a small error in your calculation.

The molarity of the sucrose solution is given as 2.40 m, which means there are 2.40 mol of sucrose per liter of solution. To find the total moles of the solution, you need to convert the molarity to moles:

2.40 mol / L * 1 L = 2.40 mol

Therefore, the mole fraction of sucrose (X) is:

moles of sucrose / total moles of solution = 2.40 mol / (2.40 mol + 55.56 mol) = 0.041

Now, you can calculate the vapor-pressure lowering:

ΔP = X * P° = 0.041 * 17.54 torr = 0.718 torr (rounded to three significant figures)

So, the vapor-pressure lowering of the sucrose solution is 0.718 torr.

(b) Aluminum Chloride (AlCl3)
Since aluminum chloride (AlCl3) is an electrolyte, it dissociates into its constituent ions when dissolved in water. According to the given assumption of 100% dissociation, each mole of AlCl3 generates three moles of particles (one mole of Al3+ ions and three moles of Cl- ions).

Taking this into account, the total moles of the solution will be:

3 moles of AlCl3 / L * 1 L = 3 moles

The mole fraction of aluminum chloride will be:

moles of AlCl3 / total moles of solution = 3 mol / (3 mol + 55.56 mol) = 0.051

Now, you can calculate the vapor-pressure lowering:

ΔP = X * P° = 0.051 * 17.54 torr = 0.895 torr (rounded to three significant figures)

So, the vapor-pressure lowering of the aluminum chloride solution is 0.895 torr.

Partly ok but not completely.

First I see where you made a typo with '
p = 0.041(17.54) and you wrote 0.41. (But that's just a typo because the 0.726 number is ok.
Second, you didn't finish the problem. You have calculated the partial pressure of the water but the problem asks for the "lowering." So you must subtract this number from 17.54 to find the lowering.
Third problem is for AlCl3. You apparently used 3 for i although it isn't labeled as such. Wouldn't i be 4 for AlCl3? Also for this you need to subtract from 17.54 to find lowering.
There is a shorcut you may use. When calculating "lowering" you can use
delta Psolvent = Xsolute*Posolvent