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**Chem** on
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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