Now, let's do this thing the right way as meticulously as possible...
Let's assume everything is at 20C. We have 10 mL of water and 3 g glucose we wish to put in solution. How much total volume will there be?
Since water has a density of 0.9982071 kg/L at 20C, and since glucose has about a 92% yield, the plato degree "P" of the solution is
3*.92/(3*.92+9.98) * 100 = 21.66
Since the relationship between plato and apparent 20C/20C gravity "g" is
P = 135.997g^3 - 630.272g^2 + 1111.14g -616.868,
we find that P = 21.66 corresponds to g = 1.0904 by root-finding techniques for this cubic equation.
Now, 1.0904 is only the apparent 20C/20C density. Since the real density of water at 20C is as I listed above, the actual density of the solution is
1.0904*.9982071 kg/L = 1.088 kg/L = 1.088 g/mL.
Since the mass of solution is 12.98 g, the volume of the solution is 12.98 / 1.088 = 11.93 mL...if you really want to get anal.
Let's assume everything is at 20C. We have 10 mL of water and 3 g glucose we wish to put in solution. How much total volume will there be?
Since water has a density of 0.9982071 kg/L at 20C, and since glucose has about a 92% yield, the plato degree "P" of the solution is
3*.92/(3*.92+9.98) * 100 = 21.66
Since the relationship between plato and apparent 20C/20C gravity "g" is
P = 135.997g^3 - 630.272g^2 + 1111.14g -616.868,
we find that P = 21.66 corresponds to g = 1.0904 by root-finding techniques for this cubic equation.
Now, 1.0904 is only the apparent 20C/20C density. Since the real density of water at 20C is as I listed above, the actual density of the solution is
1.0904*.9982071 kg/L = 1.088 kg/L = 1.088 g/mL.
Since the mass of solution is 12.98 g, the volume of the solution is 12.98 / 1.088 = 11.93 mL...if you really want to get anal.
