I can show you how to estimate it, and it will be a very close value to the true value, but I don't think this is valid if you want a more exact value. I'll explain later.
mols (NH4)2SO4 = g/molar mass = 706/132.14 = 5.343 mols.
specific volume = 0.565 mL/g
0.565 mL/g x 706 g = 398.9 mL INCREASE in volume
We have 5.343 mol/1000 g H2O. Assuming density of water @ zero C is 1.0, that is 5.343 mols/1000 mL. If we add 706 g (NH4)2 SO4 to that we now have a volume of 1000 mL + 398.9 mL = 1398.9 mL or 5.343 mols/1.3989 L = 3.8 M. The literature says it is 3.9 M for a saturated solution @ zero C. My problem with this calculation is that specific volume is mL/g. I think that means mL OF THE SOLID per g of the solid. I do not think it means that increases the volume of a solution that much. For an estimation it is OK but we know that when a solid is added to water that the volume of the SOLUTION, may decrease, increase, or stay the same. In separation of proteins by making them coagulate with saturated solutions of (NH4)2SO4 they use the procedure I've outlined above but I don't think that process gives them an accurate value for M. Close, I think, but not exact. It could be argued that specific volume DOES MEAN mL increase when added to a solution; however, every definition of specific volume I've read says that it is the reciprocal of density. So if I take 1/0.565 I get about 1.77 g/mL. That solution would be composed of 1000 g H2O + 706 g of the salt for 1706 g solution and 1706/1.77 = about 964 mL. Then 5.34 mols/0.964 L = about 5.5 M and that isn't even close to 3.9 M. So I'm thinking that (NH4)2SO4 works out that way, that makes a very easy way to get a very close estimate. Hope this helps.
The specific volume of solid ammonium sulfate is 0.565 ml/g. The solubility of ammonium sulfate at 0 oC is 706 g/1000 g water. Calculate the concentration of ammonium sulfate in a saturated solution at 0 oC.
how can i solve this question ???
my mind is blowing!!!
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Thanks So Much....
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