What mass of CsBr must be added to .500L of water to produce .4molality solution? What are mole fraction and mass percent?

Here is my work and below I will post a question about it:

1 mol CsBr = 212.81g
density of water = 1g/mL
d=mv
1=x/500
x=500g of water --> solvent

.4molality = x/.5kg
x=.2mol CsBr

1 mol CsBr = 212.81g --> .2mol = 42.6g

1 mol H20 = 18.016g
500g of H20 = 27.753mols H20

mole fraction = .2/27.953 = .00715

mass percent = 42.562/542.562 x 100 = 7.8%

MY QUESTION: The solution manual to my text book says that the mass of water is 500-mass(solute). Didn't we already establish that the mass of water is 500g by using water's density? Thank you

Your question is valid. Let's clarify the confusion.

In your calculation, you correctly determined that you have 500 grams of water as the solvent. This was determined using the density of water, which is 1 g/mL. Since you have 0.500 L of water (500 mL), the mass of the water is indeed 500 grams.

The solution manual states that the mass of water is 500 minus the mass of the solute. This is because in this context, they are assuming that the mass of the solute (in this case, CsBr) is also added to the system. So, in their approach, they consider that the mass of water is 500 - mass(solute).

Both approaches are essentially determining the mass of water in the solution, but using different perspectives. In your approach, you directly calculate the mass of water using its density. In the solution manual's approach, they assume that the mass of the solute is subtracted from the total mass (500 g) to determine the mass of water.

Both methods will yield the same result because the total mass of the solution must remain constant regardless of how the individual components are considered.

In conclusion, you have successfully calculated the mass of water as 500 grams, and you can use this value in further calculations.