posted by andy .
Calculate the volume ( L ) of the solute water and the volume ( L ) of the solvent tetrahydrofuran that should be added to generate 4.73 kg of a solution that is 1.36 m water.
m = moles/Kg solvent.
1.36 = moles water/4.73
moles H2O = 1.36 x 4.73 = 6.43
moles water amd 6.43 x 18 g/mol = 115.7 g
You can look up the density of water and the density of tetrahydrofuran and convert from grams to L. I must point out that this will NOT, howver, have a mass of 4.73 kg for the SOLUTION. The solution will have a mass of 4.73 kg + 0.116 kg = 4.85 kg.
You can go through another iteration by subtracting the water from the THF and redoing the calculation like this.
1.36 = moles H2O/(4.73-0.116)
1.36 x 4.61 = 6.28 moles water.
6.28 x 18 = 113 grams or 0.133 kg H2O.
That will be a total mass of solution of 0.113 + 4.62 kg solvent = 4.73 for the solution and that would satisfy the criteria. I don't know if you intended to write SOLUTION or solvent.
You can check this number to see if it produces 1.36 m in H2O.
113 g H2O = 6.28 moles.
That in 4.62 kg THF = 6.28/4.62 = 1.36 m