The idea here is that you need to determine how much solute you have in the concentrated solution, then figure out what volume of
Notice that the initial solution has a concentration of
This is equivalent to saying that the concentration of the initial solution must decrease by a factor of
#"D.F." = (35color(red)(cancel(color(black)(%))))/(2color(red)(cancel(color(black)(%)))) = 17.5 ->#the dilution factor
Now, you can decrease the concentration of a solution by keeping the mass of the solute constant and increasing the volume of the solution.
This means that when the concentration decreases by a factor called the dilution factor,
You thus have
#color(purple)(|bar(ul(color(white)(a/a)color(black)("D.F." = V_"diluted"/V_"initial" implies V_"diluted" = "D.F." xx V_"initial")color(white)(a/a)|)))#
In your case, the final volume of the solution must be
#V_"diluted" = 17.5 xx "50.0 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("875 mL")color(white)(a/a)|)))#
I'll leave the answer rounded to three sig figs.
So, if you increase the volume of your
There are two approaches to this calculation. The first is a simplified method which assumes that the density of the glucose solution is identical to that of water at room temperature around 1.000 g/ml.
The second method corrects for density of the glucose solution. This is more accurate especially if you preparing analytical solutions in a laboratory.
The density of 35% m/m glucose solution is 1.126 g/ml
Ive estimated the density for 2% m/m glucose around 1.007 g/ml
You can get densities for glucose solution from the internet or use a chemistry handbook that contains properties of organic compounds.
Note units are extremely important when doing calculations. Ive assumed your glucose solution is in weight percent ie m/m. This is how inorganic or organic solution densities are expressed in handbooks.
Lets start with the solution we have ie 50 ml glucose at 35% m/m
The mass of this solution can be calculated using the formula for density ie density = mass/volume
We know the density and volume. Therefore the mass of this solution is
50 ml x 1.126 g/ml = 56.3 g
Because our glucose solution is in weight percent we can easily determine the mass of glucose in solution. Ie
56.3x 35% = 19.7 g of glucose
Now the only way we can go from 35% to 2% is by dilution with distilled water. The dilute 2% glucose solution must contain the same amount of glucose.
The mass of this solution 19.7g/2%=985.3 g
Finally we use the density to determine the volume of this dilute solution = 985.3g/1.007 g/ml = 978 ml
If we used the simplified method we would obtain an answer of 875 g. Our error would have been 10.5%