I was performing some density calculation in order to double check the claimed number of points per gallon for an extract.
If one pound = 454.5 grams and one gallon = 3758 grams (at room temperature), I would expect that, once the extract is added to the water, the total weight would become 454.5 + 3758 = 4212.5 grams.
Now, if I divided 4212.5 grams/3758 mL, I would get an extract concentration of 0.121 g/mL or a density of 1.121 that three times the density reported on the books (0.04 g/mL or 1.040).
If I multiply 0.121 X 3785 I get 454.7 grams = one pound. If I do the same for 0.04 X 3785 I get 151.4 g or 0.3 pounds.
I did the experiment, by dispersing one pound of DME in one gallon of water and got a density close of 1.040.
I was thinking that, this is a matter of water buoyancy (?). The same way that you (and I) weight less inside a pool full of water, maybe the dissolved molecules weights less in water.
Please let know if I am missing something.
Thanks!, Nil
If one pound = 454.5 grams and one gallon = 3758 grams (at room temperature), I would expect that, once the extract is added to the water, the total weight would become 454.5 + 3758 = 4212.5 grams.
Now, if I divided 4212.5 grams/3758 mL, I would get an extract concentration of 0.121 g/mL or a density of 1.121 that three times the density reported on the books (0.04 g/mL or 1.040).
If I multiply 0.121 X 3785 I get 454.7 grams = one pound. If I do the same for 0.04 X 3785 I get 151.4 g or 0.3 pounds.
I did the experiment, by dispersing one pound of DME in one gallon of water and got a density close of 1.040.
I was thinking that, this is a matter of water buoyancy (?). The same way that you (and I) weight less inside a pool full of water, maybe the dissolved molecules weights less in water.
Please let know if I am missing something.
Thanks!, Nil