Okay, I'll do my best to help you out here.

1). Molality is moles of solute per kilogram of solution. The molality of choice (a) is easy since it's given as 1.0 M/Kg. For choice (b), you're given the molarity which is moles of solute per liter of solution. Instead of trying to convert the molarity to molality, you might be better off converting the molality to molarity, then going ahead and determining which is more concentrated from that point. (I.E. you should be able to easily determine the molarity of choice (a) and then use the molarity you calculated to set up a conversion factor for molarity to molality. With your conversion factor, you should be able to convert choice (b) into molality).

2). Try looking into Henry's Law.

3). In the equation you're using, the molality is not the molality of the compound, it's the molality of the ions. So in the case of choice (a), it's only 0.4 molal in total ions, while choices (b) and (c) are 0.6 molal in total ions. (Freezing point depression and boiling point depression depend on the number of ions floating around, not what they are. That is why they use CaCl

_{2} during the winter. It provides more ions than NaCl does).

4). Again, osmotic pressure depends upon number of particles floating around.

5). Once again, it's all determined by the number of ions/particles floating around, and choice (b) is 0.45 molal in terms of dissolved particles.

6). Once again, see previous question.

Questions 7, 8, and 10). Try these again, but after keeping in mind that the molality in the equations you're using is for dissolved ions/particles. (So Na

_{2}SO

_{4} would be more molal than NaCl would be).

9). Henry's Law is: S

_{gas} = K

_{H} * P

_{gas}. S

_{gas} is the solubility of the gaseous component, K

_{H} is the Henry's Law Constant, and P

_{gas} is the pressure of the dissolved gas over the solution. From the information you should be able to easily solve for K

_{H}.

11). This one's pretty easy. They tell you off the bat that the proof is two times the percentage by volume. From 92 proof we know that the solution is 46% ethanol and 54% water. To make this easy we'll do this calculation on a one liter solution. In one liter, 460 mL would be ethanol and 540 mL would be water. To calculate molarity, we need to know the mass of the ethanol so that we could then calculate the number of moles. (And since our calculations are on a one liter solution, the number of moles is our molarity). With the density of ethanol at 0.80 g/mL, we know that 460 mL weighs 368 grams. One mole of ethanol weighs 46.068 g/mol, so we can figure out that we have, to two significant figures, an 8.0 M ethanol solution.

I hope this has helped.