# Question 9176b

Jul 2, 2016

$\text{875 mL}$

#### Explanation:

The idea here is that you need to determine how much solute you have in the concentrated solution, then figure out what volume of 2% solution will contain the same amount of solute.

Notice that the initial solution has a concentration of 35% and that the diluted solution has a concentration of 2%.

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, $\text{D.F.}$, the volume increases by the same factor.

You thus have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{D.F." = V_"diluted"/V_"initial" implies V_"diluted" = "D.F." xx V_"initial}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 35% glucose solution from $\text{50.0 mL}$ to $\text{875 mL}$ you will end up with a 2%# glucose solution.

Jul 2, 2016

978 ml

#### Explanation:

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%