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