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

David--I alerted Bob Pursley to this problem and he figured out the problem. Smart man Bob Pursley.

The problem says "1.00 kg of 1.48 m CH3OH ....." That means there is a kg of the SOLUTION, not 1 kg of the solvent. Therefore, you have fewer than 1.48 moles in the kg of SOLUTION.

so would the answer be 4.8 * 10 ^ -2?? if its fewer?

To solve this problem, we need to use the formula for molality and the definition of molality.

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

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

In this case, we want to find the mass of water (solvent) that needs to be added to reduce the molality from 1.48 m to 1.00 m.

Let's assume the mass of water to be added is "x" grams.

First, calculate the moles of CH3OH in the original solution:

Moles of CH3OH = molality x mass of solvent (in kg)
= 1.48 m x 1.00 kg
= 1.48 moles

Since CH3OH is the solute, the moles of water in the original solution should be equal to the moles of CH3OH. This is based on the assumption that the volumes are additive and the density of the solution is close to the density of water.

So, Moles of water = 1.48 moles.

Now, converting the moles of water (solvent) to grams:

Mass of water (in grams) = Moles of water x molar mass of water

The molar mass of water (H2O) is approximately 18.01528 g/mol.

So, Mass of water = 1.48 moles x 18.01528 g/mol
= 26.673 g (rounded to five decimal places).

Therefore, you would need to add approximately 26.7 grams of water (rounded to two significant figures) to the 1.00 kg of 1.48 m CH3OH(aq) to reduce the molality to 1.00 m CH3OH.