How many grams of water would you add to 1.00 kg of 1.48 m CH3OH(aq) to reduce the molality to 1.00 m CH3OH. Use two sig figs.

I get 4.8 * 10^2 grams and its not right. I want to put this up here again to see if there are any errors that could have possibly been made.

(Putting this back up cause of too many questions. To Doctor Bob and Pursley)

If its actually fewer, would the answer actually be 4.8 * 10 ^-2 instead?

No.

Did Bob P respond? I don't want to redo what he has already done? If not I can get you started, perhaps right this time.

Nope he has not responded

OK. Think along these lines.

If we have 1.48 m CH3OH, that is 1.48 moles CH3OH in 1000 g water. So the solution contains 1000 g for the water + (1.48*32 = about 47.36 g CH3OH but you need to confirm that the molar mass CH3OH is 32. I just used this number off the top of my head). So the 1.48 m CH3OH solution actually has a mass of 1000 g + 47.36 g or 1047.36 grams). Now if we take a 1 kg sample of that solution, we get
1.48 moles x 1000/1047.36 = 1.41308 moles (which I know is too many places but you can always round at the last--and should). So now you want to know how much water must be added to the kg of solution (remembering that all of it is NOT water). I think something like 458.3 grams of added water will do it but you can't have that many s.f.

To solve this problem, you need to use the formula for molality:

Molality (m) = moles of solute / mass of solvent (in kg)

First, let's calculate the initial moles of solute (CH3OH) in the solution:

Molarity (M) = moles of solute / volume of solution (in L)

We have 1.48 mol CH3OH per liter of solution. Since the solution is 1.00 kg, which is equivalent to 1.00 L, the initial moles of CH3OH are:

Initial moles of CH3OH = 1.48 mol CH3OH/L * 1.00 L = 1.48 mol CH3OH

Now, we need to calculate the new mass of water needed to achieve a molality of 1.00 m.

Rearranging the molality formula:

Mass of solvent (in kg) = moles of solute / molality

Mass of solvent (in kg) = 1.48 mol CH3OH / 1.00 m CH3OH

Mass of solvent (in kg) = 1.48 kg

To find the mass of water needed, we subtract the mass of CH3OH from 1.00 kg:

Mass of water needed = 1.00 kg - 1.48 kg = -0.48 kg

It seems that there might be an error in the calculations. As we can't have a negative mass, we need to recheck the calculations.

Double-check the initial calculation of moles of CH3OH and verify whether the molarity and volume inputs are correct.

If the molarity and volume are correct, then the initial moles of CH3OH should be 1.48 mol. In that case, the correct answer would be 480 g of water, rounded to two significant figures.